Download RAY OPTICS (IMPORTANT CONCEPTS)

RAY OPTICS
(IMPORTANT CONCEPTS)
REFLECTION OF LIGHT
When ever a light ray is incident on a opaque surface it reflects back into the same media. This
phenomena is called reflection of light.
LAWS OF REFLECTION:
1.
2.
3.
4.
Point of incidence, point of normal and point of reflection lies in the same plane.
Incident ray and reflected ray lie in the same media.
The velocity, wave length and frequency of both incident ray and reflected ray are equal.
Angle of incidence is equal to angle of reflection.
i = r
REFLECTION THROUGH SPHERICAL MIRRORS:
1. Concave mirror
Focal length of concave mirror is always negative.
Images formed by concave mirror:
i)
When object is at infinite image will be formed at focus. It is real , inverted small in size.
ii)
Object is placed beyond ‘C’ image will be formed between C and F. It is real inverted
small in size compared to object.
iii)
When object is placed at C image will be formed at C only. It is real inverted in same in
size.
iv)
When object is placed between ‘C’ and ‘F’ image is formed beyond C . It is real inverted
large in size compared to object.
v)
When object is placed at focus image will be formed at infinite. It is real , inverted and
highly magnified.
vi)
An object is placed between F and P image is formed beyond the mirror. It is virtual and
vertical and large in size.
Concave mirror will give the both real and virtual images
2. Convex mirror
Focal length for convex mirror is positive.
It always leaves only virtual image.
i)
When object is infinite, image will be formed beyond the mirror. It is virtual, erect and
small in size.
ii)
When object is placed between infinite and pole P image will be formed beyond the
mirror between P and F. It is virtual, erect and small in size.
Key points:i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
ix)
x)
The radius of curvature and focal length of a concave spherical mirror is taken as
positive.
The radius of curvature and focal length of a convex mirror is taken as negative.
A real object is situated in front of mirror and therefore according to new sign
conventions, the distance of the real object from the pole of a mirror is taken as
negative.
The real image of an object is formed by a concave mirror in front of a mirror. Therefore
according to sign conventions its distance from pole of the mirror is also taken as
negative.
The virtual image of an object is formed by a concave or convex mirror beyond the
mirror. Therefore according to sign conventions its distance from pole of the mirror is
taken as positive.
The real image formed by a concave mirror is inverted and therefore according to sign
conventions its size is taken as negative.
The virtual image formed by concave or convex mirror is erect and its size is taken as
positive.
A plain mirror may be considered as a spherical mirror of infinite focal length.
Incase of a convex mirror, always virtual and diminished image is formed, irrespective of
position of object.
A plane mirror always gives virtual image of same size and same distance that of the
object.
IMPORTANT FORMULAS:
i)
ii)
R = 2f
1/f = 1/u + 1/v (Mirror formula)
iii)
iv)
v)
m= I/O = -v/u (Magnification of mirror)
m=f-v/f
m=f/f-u
NUMERICALS FOR SUPPORTIVE LEARNERS
1. A concave mirror of focal length 20cm is placed 50cm from a wall. How far from the wall an
object be placed to form its real image on the wall?
The figure shows a concave mirror at a distance of 50 cm from the wall. Suppose that object
AB is placed at distance of x from the wall to form a image A1B1on the wall. Image is formed
on same side of object.
V=-50 cm
F=-20cm
From mirror formula 1/u = 1/f – 1/v
= -1/20+ 1/50 =-3/100
U = -33.3 cm
Therefore the distance of the object from the wall x = 50 – u
X = 50 – 33.3
X = 16.7 cm.
2. An object is placed at a distance of 40cm from a concave mirror of focal
length 15cm. If the object is displaced through a distance of 20 cm towards
the mirror,
By how much distance is the image displaced?
Here f = -15 cm, u = -40 cm
Now 1/f = 1/u + 1/v
Then 1/v = 1/f – 1/u
Or
V=uf/u-f = -40 X -15/25 = -24 cm
Then object is displaced towards the mirror let u1 be the distance object from the
Mirror in its new position.
Then u1 = -(40-20) = -20cm
If the image is formed at a distance v1 from the mirror
then v1= u1f/u1-f = -20X-15/-20+15 = -60 cm.
therefore the image will move away from the concave mirror through a distance
equal to 60 – 24 = 36 cm.
3. An object is placed at distance of 25 cm from a spherical mirror and its image is formed behind
the mirror at distance of 5 cm. Find focal length? Is it concave or convex mirror?
Here u = -25 cm , v = 5 cm from the mirror formula 1/f =1/u + 1/v
Then 1/f = -1/25 + 1/5 = 4/ 25
F=6.25 cm
As the focal length is positive the mirror is convex in nature.
4. An object is placed in front of a convex mirror of radius of curvature 40 cm at a distance of 10
cm. Find the position, nature and magnification of mirror.
Here u = -10 cm, R = -40 cm
Then f = R/2 = - 20 cm
From the mirror formula 1/v = 1/f – 1/ u = -1/20 + 1/10 = 1/20.
V= 20cm so v is positive , a virtual and erect image will be formed on the other
side of the object, i.e; behind the mirror.
M=-v/u = -20/-10 = 2
5. An object is kept in front of a concave mirror of focal length of 15 cm. the
image formed is 3 times the size of the object. Calculate the two possible
distances of the object from the mirror.
Case:1. Image is real. M = -3
Here f= -15 cm
Now m=-v/u = -3
Or
V = 3u
From the mirror formula
1/f=1/u+1/v
-1/15 = 1/u + 1/3u
U=-20cm.
Case:2. When the image is virtual m = 3
Now m = -v/u = 3
Or
V=-3
From the mirror formula
1/f=1/u+1/v
Then -1/15 = 1/u-1/3u
2/3u = -1/15
U=-10cm.
HIGHER ORDER THINKING QUESTIONS:
1.
An object is kept in front of a convex mirror at a distance of 50 cm. A plane
mirror is introduced covering the lower half of the convex mirror. If the distance between the
object and plane mirror is 30 cm, it is found the there is no parallax between the images
formed by the two mirrors. What is the radius of curvature of the convex mirror?
The point object O is placed at a distance OP=50 cm from the pole P of the convex mirror M. Its
virtual image I is formed behind the mirror as shown in the figure. The plain mirror m is placed
at a distance MO=30 cm from the object, so that its distance from the pole P of the convex
mirror PM=20 cm.
Since there is no parallax between the images formed by the two mirrors, the plane mirror must
form the image at I. Since in a plane mirror, the image is as far behind the mirror as object is in
front of it.
M 1I=OM1=30cm
PI=M1I-MP = 10 cm
From the mirror formula
1/f=1/u+1/v
Here u =op = -50 cm v=PI=10 cm
1/f = -1/50+1/10 = 4/50
F= 50/4 = 12.5 cm
R=2f=25 cm
2. A container is filled with water upto high top 33.25 cm . A concave mirror is placed 15 cm
above the water level and image of an object placed at the bottom is formed 25 cm below
the water level. Find the focal length of the concave mirror? (nw=1.33)
The refractive index of water n=1.33
Height of the concave mirror above the water, x = 15 cm
Depth of the object inside the water y=33.25 cm
Depth of the image inside the water z = 25 cm
U=-(x+y/n) – (15+33.25/1.33)
U=-40 cm and v= -(x+z/n) – (15+25/1.33)
V=-33.8 cm
From the mirror formula f =uv/u+v =-40X-33.8/-40-33.8
F=-18.32 cm.
REFRACTION OF LIGHT
The phenomena of change in path of light as it goes from one media to another is
called refraction.
LAWS OF REFRACTION
i)
The incident ray, the normal to the refracting surface at the point of incidence and the
refracted ray all lie in the same plane.
ii)
The ratio of Sine of angle of incidence to the Sine of angle of refraction is constant for
the two given media
Sin i/Sin r = anb
This law is also called Snell’s law of refraction.
N=velocity of light in vacuum©/velocity of light in medium (v)
Here, n is called absolute refractive index.
KEY POINTS:-
i)
When light travels from air to medium of refractive index n, its wavelength decreases by
a factor n. i.e: becomes /n. While frequency remains same.
ii)
When white light passes through a glass slab having parallel faces, it is dispersed into its
constitutes colors also, as it undergoes refraction at the first face. However the
constituent colors recombine to produce white light, when refraction takes place at the
second face of glass slab.
iii)
For total internal reflection to take place the ray of light should travel from optically
denser to racer medium and the angle of incidence must be greater than the value of
critical angle for the pair of two media.
iv)
The knowledge of total internal reflection(TIR) is applicable in deviating a ray of light
though 900 and 1800,formation of images and blue colour of the sky etc.,
v)
The phenomena of TIR is also applicable in optical fibers which consists core coated with
low refractive index material called cladding, which are found to be of great use in the
communication, computers and etc.,,
vi)
The focal length of a lens increases when it is placed in a medium of refractive index less
than that of material of lens, and it is still behaves as the lens of same type.
vii)
In case if the lens is placed in a medium of refractive index greater than that of the
medium of lens, the nature of lens changes, i.e: convex behaves as a concave and vice
versa.
viii)
The image produced by a concave lens is always virtual and small in size.
When number of thin lenses are placed in contact a) Power of the
equivalent lens is equal to sum of the powers of the individual lenses.
b) Magnification produced is equal to the product of the magnifications produced by the
component lenses.
ix)
The inability of a lens large aperture to bring all the rays in a wide beam to
focus at a single point is called Spherical aberration.
x)
The difference in the focal length that is fp-fm is called longitudinal
spherical aberration. And the variation in the size of circular image is called “ Lateral
Spherical aberration”.
xi)
Spherical aberration can be minimized by a) Using stops b) Placing the lense so that the
total deviation is shared equally by the two surfaces of lens c) Using two plano-convex
lenses.
xii)
The ability of a lens, to converge or diverge the rays of light incident on it, is called
power of lens
IMPORTANT FORMULAS
1. n=Sin i/Sin r = C/V =0/ [absolute refractive index]
2. n12 = Sin i/Sin r = v1/v2 = 1/2
3. lateral shift D=t/Cos r(Sin (i-r))
4. n=1/Sin c where c is a critical angle
5. n2/V – n1/U = n2-n1/ R [refraction at a spherical surface]
6. Lens makers formula
1/f = (n-1)(1/R1-1/R2)
7. 1/f = 1/V – 1/U [Lens equation]
8. magnification of lens
M = I/O = V/U=f/f+u = f-v/f
9. Power of the lens P = 1/f (meter) = 100/f(cm)
10. Equivalent power p= p1+p2+p3………..
11. Equivalent magnification M = m1Xm2Xm3…………
12. If two lenses of focal length f1 and f2 are placed at a
distance d apart focal length of equivalent lens is
1/f = 1/f1+1/f2-d/f1f2
NUMERICALS FOR SUPPORTIVE LEARNERS
1. Refractive index of glass is 1.5 and that of water is 1.3. if the speed of light in water is
2.25X108m/s. What is the speed of light in glass?
Here ng=1.5 and nw=1.3 let v1 and v2 be the speeds of light in glass and water respectively. If c is
the speed of light in air then
c/v1=1.5 and c/v2=1.3
then v1/v2= 1.3/1.5
v1=1.3/1.5X2.25X108
v1=1.95X108 m/s
2. A light of wave length 6000A0 in air enters a medium with refractive index 1.5. what will be
the frequency and wave length of light in medium?
Here wave length of light in air =6000A0 = 6X10-7m
Refractive index of medium = 1.5 = n
The frequency of light does not change, when light travels from air to a refracting medium.
=c/ = 3X108/6X10-7 = 5X1014Hz
The wave length of light in the medium 1=/n = 6000/1.5 = 4000A0
3. Convex lens made up of glass of refractive index 1.5 is dipped in turn in i)
Medium A of n=1.65 ii) Medium B of n = 1.33. Explain giving reasons
Whether it will behave as a converging lens or diverging lens in each of the
Two cases.
Here ng
Let fair be the focal length of lens in air then
1/fair = (ng-1)(1/R1-1/R2)
(1/R1-1/R2) = 1/fair(ng-1) = 2/fair……………………………..(1)
i)
When lens is dipped in a medium A here nA = 1.65
Focal length be fA when dipped in a medium A then 1/fA=(nA-1)(1/R1-1/R2)
Using equation (1) we have
1/fA = (1.5/1.6 5-1) X 2/fair = -1/5.5 fair
FA = -5.5 fair
As sign of fA is opposite to that of fair, the lens will behave as diverging lense.
ii)
When lens is dipped in a Medium B nB = 1.33
Let fB be the focal length of lens when dipped in medium B
Then 1/fB = (ng-1)(1/R1-1/R2) =(ng/nB-1) )(1/R1-1/R2)
1/fB =(1.5/1.33-1)X2/fair = 0.34/1.33 fair =
FB=3.91fair as the Sign of fB is same as that of fair the lens will behave as a converging lens
4. A convergent beam of light passes through a diverging lens of focal length 0.2 meters comes
to focus at a distance 0.3 meters behind the lens find the position of the point at which the
beam would converge in the absence of lens?
F=-0.2 m, v=0.3 m
From the lens equation 1/f = 1/v-1/u
1/u = 1/v-1/f = 1/0.3-1/-0.2 = 50/6
U=6/50 = 0.12 m
In the absence of lens the beam would converge at a distance 0.12 m from the present position
of the lens.
5. A beam of light converges to a point P. A lens is placed in the path of convergent beam 12 cm
from the point P. At what point the beam converges if the lens is a) a concave lens of focal
length 16 cm b) a convex length of focal length 20 cm.
a) here u =12cm f = -16 cm the lens equation we have
V=uf/u+f=12X-16/12-16 = 48 cm as v is positive the beam converges on the same side that of
point P
b) Here u = 12 cm , f =20 cm from the lens equation we have
V=uf/u+f = 12X20/12+20 = 240/32 = 7.5 cm
As v is positive the beam converges on the same side as that of point of P
6. A converging and a diverging of equal focal lengths are placed co-axially in
Contact. Find the focal length and power of the combination.
Let f and –f are be the focal length of the converging and diverging lens
respectively then focal length of the combination
1/F = 1/f – 1/f = 0
F =  also power of the combination p = 1/F = 0.
HIGHER ORDER THINKING QUESTIONS.
1. A liquid of refractive index 1.5 is poured into cylindrical jar of radius 20cm up to height of 20
cm. A small bulb is lighted at the center of the bottom of the jar. Find the area of liquid
surface thorough which the light of the bulb passes into air.
The point L shows the position of bulb at the bottom of a jar containing liquid (n=1.5) up to the
height h =20cm. the light will emerge out through the larges area it is through a circle of
maximum radius r, if a refraction at liquid-air interface, it comes out parallel to the surface .
It will happen so if the angle of the incidence for the rays like incident at the points on the
circumference of the circle of radius r is equal to the critical angle for liquid-air interface. If c is a
critical angle then
N = 1/Sin c
Sin c = 1/n = 1/1.5 = 0.6667
C = 41.80 from right angled triangle AOL we have Tan c = OA/OL = r/h
R =h tan c
R=20 X tan 41.80= 17.882 cm
Therefore required area =  r 2
= 3.14 X (17.882)2
=1004.6cm2
2. Explain what happens when a convex lens of refractive index 1.2 is immersed in a liquid of
refractive index 1.3 .
Let f be the focal length of the lens , then
1/f =(n-1) 1/R1-1/R2
=(1.2-1) 1/R1-1/R2 = 0.2 X 1/R1-1/R2
1/R1-1/R2= 1/0.2f
If f1 is the focal length of lens , when immersed in a liquid then
1/f 1=(n1-1)(1/R1-1/R2)
Where n1 = 1.2/1.3 is the refractive index of material of lens with respective liquid.
1/f1 = (1.2/1.3-1)(1/R1-1/R2) = -1/13 (1/R1-1/R2)
1/f1= - 1/13 X 1/0.2 f
F1 = -2.6f the convex lens act as a concave lens.
DISPERSION
KEY POINTS:-
1. There are two values of angle of incidence for which a ray passing through a prism
deviates through a given angle.
2. When a prism made of material of refractive index n is placed in a medium of
refractive index n1(n1< n). the prism still deviates the light towards its base. But
angle of deviation decreases.
3. a prism disperses white light into constant colours due to the fact that the light of
different wave lengths undergo unequal deviations , while passing through the
prism. It is because; refractive index of material of prism depends upon the wave
length of light.
4. Refractive index of a material of yellow light (n is practically equal to the mean of
the refractive indexes for the violet and red lights. For this reason yellow light is
called mean light.
5. The angular dispersion for any two colours may be defined as the difference in
deviations suffered by the two colours in passing through the prism.[].
6. The dispersive power of the material for any two colours may be defined as the
ratio of the angular dispersion for these two colours to the deviation suffered by
mean light. It is denoted by .
7. The intensity of light corresponding to a wave length in the scattered light varies
inversely as the fourth power of the wave length. If  is a wave length in scattered
light , then the amount of scattering  1/4
it is called Rayleigh’ss law of scattering.
8. In the primary rainbow, the upper edge is red, while in case of secondary rainbow
upper edge is violet.
IMPORTANT FORMULAE:
1.
2.
3.
4.
5.
6.
7.
8.
Refractive index of prism n = Sin(A+m)/Sin A/2)
at the minimum deviation I = A +m/2
A = r1+r2
Deviation  = A(n-1)
ny = nv+nr/2
Angular dispersion  = v - r = A (nv – nr)
Dispersive power  = v - r/ = (nv – nr) / n-1
NUMERICALS FOR SUPPORTIVE LEARNERS
1. A ray of light incident at 490 on the face of an equilateral prism passes symmetrically calculate
the refractive index of the material of the prism?
As the prism is an equilateral one A = 600. Further a ray of light incident at 490 passes
symmetrically through the prism, it is in the minimum deviation position we know that when the
prism is placed in minimum deviation position.
R=A/2 = 60/2=300 and i=490
N=Sin i/Sin r = Sin 490/ Sin 300 = 1.51
2. A ray of light passes through an equilateral glass prism, such that angle of
incidence is equal to the angle of emergence. If the angle of emergence is ¾ times the angle of
prism, calculate the refractive index of glass prism?
As the prism is an equilateral one A = 600
Also I = e = ¾ A = ¾ X 60 0 = 45 0
Further as the angle of incidence is equal to the angel of emergence, the prism is in minimum
deviation position. We know that when the prism is placed in minimum deviation position r =
A/2 = 300 and I = 450
N = Sin i/Sin r = 1/2 X 2 =2.
3. A ray of light is incident normally on one of the faces of the prism of apex
angle 300 and refractive index 2 . find the angle of deviation for the ray of
light?
When the ray of light PQ is instant normally on the face AB of the prism ABC
angle A = 300, it does not get refracted . it travels state along the path QR and
comes to incident on the face AC of the prism at an angle I = 300.
If r is the angle of refraction, then
1/n = Sin i/Sin r
1/2 =Sin 300/Sin r
Sin r = 1/2
R=450.
The angle of deviation  = r-I = 450-300 = 150
 =150
4. A ray of light incident on the face AB of a glass prism ABC having the vertex
angle A = 400. the face AC is silvered and a ray of light is incident on the face AB such that it
retraces its path. If refractive index of the glass is 1.54 calclate the angle of incidence on the
face AB?
Here A = 400,N=1.54
Suppose that the ray of light PQ is incident on the face AB at an angle I, such that
The angle of refraction is r, the ray will retrace its path, if the refracted ray QR falls on the silver
surface normally as shown.
Since ∆ ARQ is a right angled triangle, it follows that r1=400
Now n=Sin i/Sin r
Sin i=nXSin r
=1.54 X 0.6428 = 0.9899
I=810511
5. Prism of refractive index 1.53 is placed in water of refractive index 1.33. if
the angle of prism is 600, calculate the angle of minimum deviation in water?
Here A = 600, ng = 1.53, nw= 1.33
Refractive index of material of prism with respect to water
Nwg=ng/nw = 1.53/1.33 = 1.15
Let m be the angle of minimum deviation on placing the prism in water then
Nwg = Sin (A+m)/2)/Sin A/2
1.15 = Sin (60+m)/2)/Sin 60/2
Sin (60+m)/2 = 1.15 X Sin 300 = 0.575
60+m)/2=35.10
=m = 10.20
OPTICAL INSTRUMENTS
Key points:1. Eye defects are four types.
i)
Myopia
ii)
Hypermetropia
iii)
Presbyopia
iv)
Astigmatism
2. Magnifying power of simple microscope is
M = 1+D/f
When object is at infinite
M = D/f
3. A magnifying power of compound microscope
M = v/u [1+D/fe]
M = -L/fo[1+D/fe]
M = v/u X D/fe
4. Magnifying power of Astronomical telescope in normal adjustment
M = -fo/fe
When image is formed at least distinct vision
M =--fo/fe[1+fe/D]
5. The reciprocal of smallest distance between two objects which are
clearly distinguished by microscope is called resolving power of
microscope.
R.P=1/d = 2n Sin/ . Where d is the distance between the objects,
N is the refractive index of the objective lens.
 is the angle of cone.
6. The reciprocal of smallest angular separation at the objective lens made
by the light rays which are coming from infinite object is called
resolving power of telescope.
R.P = 1/d = a/1.22 . Where d angular separation
 Wave length of light. A aperture of objective lens.
1. A certain person can see clearly objects at a distance between 20 cm and 250 cm from his eye.
i) What spectacles are required to enable him to see distant objects clearly? ii) What will be
his least distance of distinct vision, when he is wearing spectacles?
i)
To see the distant objects
The distance of the far point, x = 250 cm = 2.5 meters
The defect can be correct by using concave lens of focal length f = -x = - 2.5 meters
The power of the lens is given by
P=1/f[m] = -1/2.5=-0.4 D.
ii)
Here f =-250 cm at the distance of near point x = 20 cm
Let D with a least distant of distinct vision, when the person is wearing the
spectacles. Now f = xD/x-D = -250 = 20 X D/20-D
-250[20-D] = 20D
-5000 + 250 D = 20D
230 D = 5000
D = 18.52 cm
2.
A defectiveness a converging of focal length 12 cm to examine the fine details
of some fibers found at the sense of a crime.
i)
ii)
i)
What is the maximum magnification given by the lens?
What is the magnification of relaxed eye viewing?
Here f =12 cm, the maximum magnification occurs, when the image is found by the lens
at least distance of distinct vision that is at D = 25 cm
Therefore maximum magnification given by the lens.
M = 1 +D/f=1+25/12= 3.1
ii)
The eye is most relaxed when the object lies at infinite therefore magnification for
relaxed eye viewing is
M=D/f = 25/12=2.1
2. A converging lens of focal length 6.25 cm is used as a magnifying glass. If the near point of the
observer is 25 cm from the eye and the lens is held close to the eye. Calculate the distance of
object from the lens and angular magnification. Also find angular magnification when the final
image at infinite?
Here f = 6.25 cm , v = -D = -25 cm
Let u be the distance of object from the lens then
1/f = 1/v-1/u
1/6.25 = -1/25-1/u then
U=-5 cm.
Now angular magnification M =1+D/f
M=1+25/6.25 = 5.
But , when the final image is formed at infinite. The angular magnification is
given by M = D/f = 25/6.25 = 4.
4.
A compound microscope has magnification of 30. the focal length its eye piece is 5 cm
assuming that the final image to be formed at least distance of distinct vision. Calculate the
magnification produced by the objective.
M=-30[Magnifying power is negative]
Fe=5 cm , D=25 cm
M = moXme = mo X[1+D/fe]
-30 = mo[1+25/5]
Mo =-30/6 =-5.
5. The focal lengths of the objective and eye-piece of a compound microscope are 4 cm and 6 cm
respectively. If an objective is placed at distance of 6 cm from the objective, what is the
magnification produced by the microscope?
Here fo=4 cm , fe=6 cm, D=25 cm, uo=-6 cm
Suppose that the image of the object is formed by the object lens at distance v 0 from it. The
distance v0 is given by
1/fo= 1/v0-1/u0
1/v0=1/f0+1/u0
=1/4-1/6=2/24
V0=12cm
Magnification m=v0/u0 (1+D/fe)
=-12/6[1+25/6]
=-10.32.
6. The magnifying power of an astronomical telescope in the normal adjustment position is 100.
The distance between the objective and eye-piece is 101 cm. calculate the focal lengths of the
objective and eye-piece?
Here fo+fe=100cm
M=-100
M=f0/fe
100= f0/fe
Fo=100 fe
100 fe+fe=101
Then fe=1 cm
And fo=100 cm
7. A reflecting type telescope as a concave reflector of radius of curvature 20 cm. calculate the
focal length of eye-piece to secure the magnification of 20?
Here M=-20
Radius of curvature of concave reflector
R=-120 cm
Therefore focal length of the concaved reflector
Fo=R/2 = -60cm
M= f0/fe
Fe=fo/M=-60/-20=3 cm
And fe=3 cm.
8. The magnifying power of an astronomical telescope is 5. when it is set for normal adjustment,
the distance between the two lenses is 24 cm. Calculate the focal length of eye-piece and that
of its objective?
Here M =-5
fo+fe=24cm
M = f0/fe
Fe=5fe
5fe+fe=24 cm
6fe=24 cm
Fe=4 cm
Fo=20 cm
9. A converging lens of power 25D is used as simple microscope. Calculate the magnifying power
if D = 25 cm?
P=25D
1/f=P=25D
F=4cm
Magnifying power M=1+D/M=1+25/4=1+6.25
M=7.25
10. The total magnification produced by a compound microscope is 20,while that produced by the
eye-piece alone is 5. when the microscope is focused on a certain object the distance between
the objective and eye-piece 14 cm . find focal length of objective and eye-piece? D=20cm
M=-20
Magnification produced by the eye-piece
Me=5
But M=moXme
Mo=-20/5=-4
But for compound microscope we have
Me=1+D/fe
5=1+20/fe
Fe=5cm
Me=fe/ue+fe
5=5/ue+5
Ue=-5 cm
Distance between objective and eye-piece
Ve+ue=14
Vo=100cm
Mo=vo/uo
-4=10/uo
Uo=-2.5 cm
Then 1/fo=1/vo-1/uo
1/fo=1/10+1/2.5=5/10
Fo=2cm.
RAY OPTICS
Question 9.1:A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At
what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of
the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Answer
Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = −27 cm
Radius of curvature of the concave mirror, R = −36 cm
Focal length of the concave mirror,
Image distance = v
The image distance can be obtained using the mirror formula:
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain
the image.
Question 9.2:
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the
magnification. Describe what happens as the needle is moved farther from the mirror.

Answer
Height of the needle, h1 = 4.5 cm
Object distance, u = −12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
Hence, magnification of the image,
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image
will reduce gradually.
Question 9.3:
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is
measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive
index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Answer
Actual depth of the needle in water, h1 = 12.5 cm
Apparent depth of the needle in water, h2 = 9.4 cm
Refractive index of water = μ
The value of μcan be obtained as follows:
Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index,
The actual depth of the needle remains the same, but its apparent depth changes. Let y be the new apparent depth of the
needle. Hence, we can write the relation:
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the
microscope should be moved up.
Distance by which the microscope should be moved up = 9.4 − 7.67
= 1.73 cm
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of
water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a
point source.)

Answer
Actual depth of the bulb in water, d1 = 80 cm = 0.8 m
Refractive index of water,
The given situation is shown in the following figure:
Where,
i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of radius,
Using Snell’ law, we can write the relation for the refractive index of water as:
Using the given figure, we have the relation:
R = tan 48.75° × 0.8 = 0.91 m
Area of the surface of water = πR2 = π (0.91)2 = 2.61 m2
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m2.
Question 9.6:
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle
of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting
angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum
deviation of a parallel beam of light.

Answer
Angle of minimum deviation,
= 40°
Angle of the prism, A = 60°
Refractive index of water, µ = 1.33
Refractive index of the material of the prism =
The angle of deviation is related to refractive index
as:
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let
be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is given by the relation:
Hence, the new minimum angle of deviation is 10.32°.
Question 9.7:
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of
curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Answer
Refractive index of glass,
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R1
Radius of curvature of the other face of the lens = R2
Radius of curvature of the double-convex lens = R
The value of R can be calculated as:
Hence, the radius of curvature of the double-convex lens is 22 cm.
Question 9.7:
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of
curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Answer
Refractive index of glass,
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R1
Radius of curvature of the other face of the lens = R2
Radius of curvature of the double-convex lens = R
The value of R can be calculated as:
Hence, the radius of curvature of the double-convex lens is 22 cm.
Question 9.9:
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by
the lens. What happens if the object is moved further away from the lens?

Answer
Size of the object, h1 = 3 cm
Object distance, u = −14 cm
Focal length of the concave lens, f = −21 cm
Image distance = v
According to the lens formula, we have the relation:
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is
erect and virtual.
The magnification of the image is given as:
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not
beyond it. The size of the image will decrease with the increase in the object distance.
Question 9.10:
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the
system a converging or a diverging lens? Ignore thickness of the lenses.

Answer
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Focal length of the system of lenses = f
The equivalent focal length of a system of two lenses in contact is given as:
Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as
a diverging lens.
Question 9.11:
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm
separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image
at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in
each case?

Answer
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision,
Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Magnitude of the object distance,
= 2.5 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
Image distance for the eyepiece,
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
Image distance for the objective lens,
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Magnitude of the object distance,
= 2.59 cm
The magnifying power of a compound microscope is given by the relation:
Hence, the magnifying power of the microscope is 13.51.
Question 9.12:
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an
eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the
separation between the two lenses? Calculate the magnifying power of the microscope,

Answer
Focal length of the objective lens, fo = 8 mm = 0.8 cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objective lens, uo = −9.0 mm = −0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = −d = −25 cm
Object distance for the eyepiece =
Using the lens formula, we can obtain the value of
as:
We can also obtain the value of the image distance for the objective lens
using the lens formula.
The distance between the objective lens and the eyepiece
The magnifying power of the microscope is calculated as:
Hence, the magnifying power of the microscope is 88.
Question 9.13:
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the
magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answer
Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
The separation between the objective lens and the eyepiece is calculated as:
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150
cm.
Question 9.14:
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length
1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective
lens? The diameter of the moon is 3.48 × 106 m, and the radius of lunar orbit is 3.8 × 108 m.

Answer
Focal length of the objective lens, fo = 15 m = 15 × 102 cm
Focal length of the eyepiece, fe = 1.0 cm
(a) The angular magnification of a telescope is given as:
Hence, the angular magnification of the given refracting telescope is 1500.
(b) Diameter of the moon, d = 3.48 × 106 m
Radius of the lunar orbit, r0 = 3.8 × 108 m
Let
be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm
Question 9.15:
Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the
pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of
images that one obtains from explicit ray diagrams.]

Answer
(a) For a concave mirror, the focal length (f) is negative.
f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
u < 0
For image distance v, we can write the lens formula as:
The object lies between f and 2f.
Using equation (1), we get:

is negative, i.e., v is negative.
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
f>0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
u<0
For image distance v, we have the mirror formula:
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
(c) For a convex mirror, the focal length (f) is positive.
f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative,
u < 0
For image distance v, we have the mirror formula:
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
u < 0
It is placed between the focus (f) and the pole.
For image distance v, we have the mirror formula:
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:
Magnification, m
>1
Hence, the formed image is enlarged.
Question 9.16:
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to
be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of
glass = 1.5. Does the answer depend on the location of the slab?

Answer
Actual depth of the pin, d = 15 cm
Apparent dept of the pin =
Refractive index of glass,
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
The distance at which the pin appears to be raised =
For a small angle of incidence, this distance does not depend upon the location of the slab.
Question 9.17:
(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of
the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of
the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?

Answer
(a) Refractive index of the glass fibre,
Refractive index of the outer covering of the pipe,
= 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the inner core − outer core interface is given as:
For the critical angle, total internal reflection (TIR) takes place only when
Maximum angle of reflection,
, i.e., i > 59°
Let,
be the maximum angle of incidence.
The refractive index at the air − glass interface,
We have the relation for the maximum angles of incidence and reflection as:
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe,
For the angle of incidence i = 90°, we can write Snell’s law at the air − pipe interface as:
.
Question 9.18:
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under
some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen.
Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a
contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or
shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or
decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond
cutter?

Answer
(a) Yes
Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a
point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a
real image will be formed.
(b) No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at
the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
(c) The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that
the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer
medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
(d) Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one
medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.
(e) Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than
that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally
reflected from its faces. This is the reason for the sparkling effect of a diamond.
Question 9.19:
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of
a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer
Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens =
For real images, the maximum focal length is given as:
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Question 9.20:
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different
locations separated by 20 cm. Determine the focal length of the lens.

Answer
Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
Therefore, the focal length of the convex lens is 21.39 cm.
Question 9.21:
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm
apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel
light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between
the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of
the image.

Answer
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d = 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
Where,
= Object distance = ∞
v1 = Image distance
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
Where,
= Object distance
= (30 − d) = 30 − 8 = 22 cm
= Image distance
The parallel incident beam appears to diverge from a point that is
the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
from the centre of
Where,
= Object distance = −∞
= Image distance
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
Where,
= Object distance
= −(20 + d) = −(20 + 8) = −28 cm
= Image distance
Hence, the parallel incident beam appear to diverge from a point that is (420 − 4) 416 cm from the left of the centre of the
combination of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of
effective focal length does not seem to be useful for this combination.
(b) Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens,
According to the lens formula:
Where,
= Image distance
Magnification,
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:
Where,
= Object distance
= +(120 − 8) = 112 cm.
= Image distance
Magnification,
Hence, the magnification due to the concave lens is
.
The magnification produced by the combination of the two lenses is calculated as:
The magnification of the combination is given as:
Where,
h1 = Object size = 1.5 cm
h2 = Size of the image
Hence, the height of the image is 0.98 cm.
Question 9.22:
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total
internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer
The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.
Angle of prism, A = 60°
Refractive index of the prism, µ = 1.524
= Incident angle
= Refracted angle
= Angle of incidence at the face AC
e = Emergent angle = 90°
According to Snell’s law, for face AC, we can have:
It is clear from the figure that angle
According to Snell’s law, we have the relation:
Hence, the angle of incidence is 29.75°.
Question 9.24:
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The
cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens
behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of
converging power of the eye-lens) of a normal eye.

Answer
Least distance of distinct vision, d = 25 cm
Far point of a normal eye,
Converging power of the cornea,
Least converging power of the eye-lens,
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D
Power of the eye-lens is given as:
To focus an object at the near point, object distance (u) = −d = −25 cm
Focal length of the eye-lens = Distance between the cornea and the retina
= Image distance
Hence, image distance,
According to the lens formula, we can write:
Where,
= Focal length
Power of the eye-lens = 64 − 40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.
Question 9.25:
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its
ability of accommodation? If not, what might cause these defects of vision?

Answer
A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs
when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the
eye-lens loses its ability of accommodation, the defect is called presbyopia.
Question 9.26:
A myopic person has been using spectacles of power −1.0 dioptre for distant vision. During old age he also needs to use
separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Answer
The power of the spectacles used by the myopic person, P = −1.0 D
Focal length of the spectacles,
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles,
the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to
see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power,
The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the
objects placed at 25 cm.
Question 9.27:
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the
vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer
In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the
refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called
astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane
is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This
defect can be corrected by using cylindrical lenses.
Question 9.28:
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal
length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the
book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple
microscope?

Answer
(a) Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm
Closest object distance = u
Image distance, v = −d = −25 cm
According to the lens formula, we have:
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u’), the image distance
According to the lens formula, we have:
Hence, the farthest distance at which the person can read the book is
5 cm.
(b) Maximum angular magnification is given by the relation:
Minimum angular magnification is given by the relation:
Question 9.29:
A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a
converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.

Answer
(a) Area of each square, A = 1 mm2
Object distance, u = −9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
Magnification,
Area of each square in the virtual image = (10)2A
= 102 × 1 = 100 mm2
= 1 cm2
(b) Magnifying power of the lens
(c) The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is
and the magnifying power is
.
The two quantities will be equal when the image is formed at the near point (25 cm).
Question 9.30:
(a) At what distance should the lens be held from the figure in
Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case?
Explain.

Answer
(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).
Image distance, v = −d = −25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have:
Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.
(b) Magnification =
(c) Magnifying power =
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
Question 9.31:
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each
square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very
close to the magnifier?
[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and
the angular magnification (or magnifying power) of an instrument.]

Answer
Area of the virtual image of each square, A = 6.25 mm2
Area of each square, A0 = 1 mm2
Hence, the linear magnification of the object can be calculated as:
Focal length of the magnifying glass, f = 10 cm
According to the lens formula, we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence,
it cannot be seen by the eyes distinctly
Question 9.32:
Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced
by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular
magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us
from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short
distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Answer
(a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the
object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A
closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the
object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the
angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle
subtended at the lens. Image distance does not have any effect on angular magnification.
(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very
small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small
focal length.
(d) The angular magnification produced by the eyepiece of a compound microscope is
Where,
fe = Focal length of the eyepiece
It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as
Where,
= Object distance for the objective lens
= Focal length of the objective
The magnification is large when
>
. In the case of a microscope, the object is kept close to the objective lens.
Hence, the object distance is very little. Since
the given condition.
is small,
will be even smaller. Therefore,
and
are both small in
(e)When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted
light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The
precise location of the eye depends on the separation between the objective lens and the eyepiece.
Question 9.33:
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece
of focal length 5 cm. How will you set up the compound microscope?

Answer
Focal length of the objective lens,
= 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
The angular magnification of the objective lens (mo) is related to me as:
=m
Applying the lens formula for the objective lens:
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
Where,
= Image distance for the eyepiece = −d = −25 cm
= Object distance for the eyepiece
Separation between the objective lens and the eyepiece
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Question 9.34:
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the
magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image
is at infinity)?
(b) the final image is formed at the least distance of distinct vision
(25 cm)?

Answer
Focal length of the objective lens,
= 140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
(a) When the telescope is in normal adjustment, its magnifying power is given as:
(b) When the final image is formed at d,the magnifying power of the telescope is given as:
Question 9.35:
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by
the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?

Answer
Focal length of the objective lens, fo = 140 cm
Focal length of the eyepiece, fe = 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece
(b) Height of the tower, h1 = 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m
The angle subtended by the tower at the telescope is given as:
The angle subtended by the image produced by the objective lens is given as:
Where,
h2 = Height of the image of the tower formed by the objective lens
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
(c) Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation:
Height of the final image
Hence, the height of the final image of the tower is 28.2 cm.
Question 9.36:
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If
the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an
object at infinity be?

Answer
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R1 = 220 mm
Hence, focal length of the objective mirror,
Radius of curvature of the secondary mirror, R1 = 140 mm
Hence, focal length of the secondary mirror,
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary
mirror.
Hence, the virtual object distance for the secondary mirror,
Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:
Hence, the final image will be formed 315 mm away from the secondary mirror.
Question 9.37:
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A
current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a
screen placed 1.5 m away?

Answer
Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ= 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
Hence, the displacement of the reflected spot of light is 18.4 cm.
Question 9.38:
Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A
small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the
needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is
repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Answer
Focal length of the convex lens, f1 = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
Let the refractive index of the lens be
of the other surface is −R.
and the radius of curvature of one surface be R. Hence, the radius of curvature
R can be obtained using the relation:
Let
be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror =
Radius of curvature of the liquid on the side of the lens, R = −30 cm
The value of
can be calculated using the relation:
Hence, the refractive index of the liquid is 1.33.
Ray optic (Additional Questions)
Very short answer type question
1. Name the Principle on which an optical fibre works.
2. Which of the two main parts of an optical fibre has a higher value of refractive index?
3. When a Ray of light is passed through a Prism, under what condition will the angle of Incidence
be equal to the angle of Emergence?
4. An object is placed at the focus of a concave Lens. Where will its image be formed?
5. A lense of Glass is immersed in water. What will be its effect on the power of the lens.
6. A convex lens made of Glass of refravtive index µL is immerse in a medium of refractive index
µm. How will the lens behave when µL less than µ M?
7. How does the resolving of power of a Telescope Change, if the incident yellow light is replaced
by blue light.
8. The refractive index of a medium is √3. What is the angle of refraction, if the unpolarised light is
incident on it at the polarising angle of the medium
9. What is focal length of a plane mirror?
10. Two thin lenses of power + 7D and – 3 D are in contact. What is the focal length of the
combination.
11. A thin prism of 600 angle gives a deviation of 30. What is the refractive index material of Prism?
12. A concave Lens is Placed in water will there be any change in focal length? Give reason.
13. Why does the sky appear blue?
14. What is resolving power of microscope?
15. What is the relation between resolving power and limit of resolution of an optical instrument.
16. Define critical angle for total internal reflection.
17. Write the value of critical angle for a material of refractive index √3.
18. What is an optical fibre.
19. What is the main use of optical fibre.
20. Name the physical Principal on which the working of a optical fibre is based.
21. State Rayleigh’s law of Scattering?
22. What is the distance between the objective and the eye piece of a Telescope in normal
adjustment.
23. What is the advantage of Galileo’s Telescope over Terrestrial Telescope.
24. How does the Power of a Convex lense vary, if the incident red light is replaced by violet light.
25. A double convex lens made from a material of refractive index µ 1, is immerse in a liquid of
refractive index µ 2, where µ2> µ1. What Change, if any, would occure in the nature of the lens
Short answer type question:(Two or three Marks)
1. Draw a labeled Ray Diagram of reflecting type Telescope.
2. A concave lens has the same radi of curvature for both side and has a refractive index 1.6 in
year. In the second case it is immerse in a liquid of refractive index 1.4. Calculate the ratio of
the focal length of the lens in the two cases.
3. Draw a lebel Ray diagram to show the image formation in refracting type Telescope. Why
should the Diameter of the objective of Telescope be large.
4. Give reason for the following ; the sun looks radish at sun rise at sun set as viewed from the
earth.
5. Explain how does the refraction of light affect the length of the day.
6. Derive the rationship between the refractive index and critical angle for a given pair of
media.
7. Write two conditions necessary for total internal reflection to take place.
8. Monochromatic light is refracted from air into glass of R.I. n. find the ratio of wave length of
the incident and refracted light.
9. A right angled crown glass prism with critical angle 410 is placed before an object PQ, in two
positions as shown in figure I and II. Trace the path of rays from P and Q Passing through the
Prisim in th two cases.
P
↑
↑
Q
10. A converging and a diverging lens of equal focal lens are placed co- axially in contact. Find
the power and focal length of the combination.
11. Using the lense formula show that an object placed between the optical Centre and the
focus of a convex lens produces a virtual and enlarge image.
12. Draw a labelled ray diagram showing the formation of an image of an distant object using an
astronomical Telescope in the normal adustment position.
13. Draw a ray diagram of a compound mirocsope.
14. Draw a ray diagram of a simple microscope.
15. Give reason (1).sun is visible before the actual sun rise(.2). sun looks radish at sun set.
16. What changes in the focal lenghth of a concave mirror and a convex lens occur , when the
incident violet light is replaced with white light.
17. State the condition for Toatl internal reflection of llight to take place. Calculate the the
speed of light in a medium whose critcal angle is 450.
18. How will the resolving power of a compound microscope be affected when the frequency of
light used to illuminate the object is increased and the focal length of the objective lens is
increased? Justify your answer your answer in each case.
19. Draw a lebelled ray diagram showing the formatioin of image using a Newtonian type
reflecting Telescope.
20. State the reason for the following observation recored from the surface of the moon.
I) Sky appears dark
II) Rainbow is never formed.
21. How will be the focal length of a lens be affected if it is deeped in water.
22. Define resolving power of Telescope and microscope. Write the mathematical formaula.
23. Three rays of light – red(R), Green ( G ) and Blue (B)- are incident on the face A B of a right
angled Prism ABC. The refractive indexes of the material of the Prism for red, Green Blue
wave lengths are 1.39, 1.44 and 1.47 respectively. Trace the path of rays through the prism.
How will the situatiin change if these rays were incident normally on one of the faces often
equilateral Prism?
B→
G→
R→
24.
What do you understand by the phenomenon of the total internal
releflection? Give two condition for its to take place.
25.
Derive the relation connectiong refractive index and critical angle for a
given pair of media for toal internal reflection.
Long answer type question
1.
2.
3.
4.
5.
Derive the expression for refractive index of the material of the Prism in
terms of angle of Prism and angle of minimum deviation. Use this formula to calculate the
angle of minimum deviation for an equilateral triangular prism of refactive index √3.
Draw a label ray diagram of an astonomical Telescope for the near point
adjustment. You are given three lenses of power 0 .5D, 4D, 10D. State with reason, which
two lenses will you select for constructing for good astronomical Telescope. Calcualte the
resolving power of this Telescope assuming the diameter of the objective lense to be 6 cm
and wave length of light used to be 540nm.
Draw a label ray diagram of a compound microscope showing the
formatin of image at the near point of the eye.
A compound microscope uses an objective lense of focal length of 4 cm and eye
lens of 10 cm. An object is placed at 6cm from the objective lens. Calculate the magnifying
power if the final image is formed at near point. Also calculate length of the compound
microscope.
Draw a label ray diagram of astronomical Telescope forming the image at
infinity. An astronomical Telescope uses two lenses of power 10D, 1D. State with reason
which lens is prefered as objective as eye piece. Calculate the magnifying colour of the
Telescope if the final image is at near point. How do the light gathering power of a
Telescoper change if the aparture of the objective lense is doubled.
Derive the lens makers formula incase of a double convex lens. Sate the
assumtion made and the sign convention used.
6.
With the help of ray diagram, show the formation of image of a point
object by refraction of light at spherical surface separating two media of refrative index n1
and n2(n2 >n1) respectively. Using this diagram derive the relation.
.
7.
How do you estimate rough focal length of a convex lens. Draw a ray
diagram to show image formation by diverging lens. Using this diagram, derive the relation
between object distanc u, image distance v and focal length F of the lens. Sketch the graph
between
and 1/v for this lens.
8.
Two lenses of power + 15 D and -5D are in contact with each other
forming a combination lens. what is the focal length of this combination? An object of size
3cm is placed at 30 cm from this combination of lennes. Calculate the position and the size
of the image formed.
9.
Draw a label diagram of a simple microscope and show that its
magnification m is given by
M= 1+D/F where F= focal length and D= least distance of distinct vision.