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XI SCIENCE
[PHYSICS – II]
RAY OPTICS
XI SCIENCE NOTES
Prof. R. S. Gade.
February 3, 2013
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 1
RAY OPTICS
[SYLLABUS]
1.
2.
3.
4.
5.
6.
7.
8.
Reflection of light by spherical mirrors
Refraction at single curved surface: Derivation is expected.
Lens maker‟s equation: Derivation of equation for an ideal lens.
Concept of conjugate foci: Concept only.
Magnifying power of simple microscope: Definition, derivation of expression.
Magnifying power of compound microscope: Definition, derivation of expression.
Magnifying power of telescope: Definition, derivation of expression.
Lens defects: Physical interpretation of spherical and chromatic aberration only.
* * * * *
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 2
RAY OPTICS
Reflection of light by spherical mirrors:
Spherical mirror is spherical reflecting surface which may be taken to be a section of
a hollow sphere. There are two types of the spherical mirrors as – (i) a concave mirror,
which has its inner surface reflecting and (ii) a convex mirror, which has its outer
surface reflecting.
For any mirror we can define following terms:
1. Pole: The centre of a spherical mirror or surface is called the pole of the surface or
spherical mirror or optical centre of the mirror.
2. Centre of curvature: The centre of the sphere of which the spherical surface or mirror
is a section is called the centre of curvature of the spherical mirror or surface.
3. Radius of curvature: The radius of the sphere of which the spherical mirror or surface
is a section is called the radius of curvature of the spherical mirror or surface.
4. Principal axis: The line passing through the pole and the centre of curvature of a
spherical mirror or surface is called the principal axis of the spherical mirror or surface.
5. Principal focal point: A narrow beam of light parallel to the principal axis after
reflection from a spherical mirror either converges to (for a concave mirror) or appear to
diverse from (for a convex mirror) a single point on the principal axis. This point is
called the principal focal point or principal focus of the spherical mirror.
6. Focal plane: A narrow parallel beam of light close to but not parallel to the principal
axis, converges at (or appears to diverge from) a single point on a plane perpendicular
to the principal axis and passing through the principal focus. Such a plane is called the
focal plane.
Laws of reflection:
When the light is reflected from the spherical mirror, it obeys the following laws of
reflections:
1. The incident and reflected rays and the normal to the reflecting surface at the point of
incidence, lie in the same plane.
2. The incident and reflected rays make equal angles with the normal on opposite sides of
it.
New sign conventions:
1. When drawing the ray diagram the object must be at the left of the lens.
2. All distances are measured from optical centre of lens (or pole of the mirror) as origin.
3. Distances measured against the direction of the incident ray are negative.
4. Distances measured in the direction of the incident ray are taken as positive.
5. Distances measured in the upward from the axis are taken as positive while as in the
downward direction are taken as negative.
According to sign conventions the radius of curvature for the convex lens is positive
while that of concave lens is negative.
Image formation by concave mirror:
(1) Object at infinity: A real image is formed in the focal plane. It is inverted and highly
diminished relative to the object.
(2) Object between infinity and centre of curvature: A real image is formed between the
focal plane and centre of curvature. It is inverted and diminished relative to object.
(3) Object at the centre of curvature: A real image is formed at the centre of curvature. It
is inverted relative to the object and of the same size as that of the object.
(4) Object between the centre of curvature and focal plane: A real image is formed
beyond the centre of curvature. It is inverted and magnified relative to the object.
(5) Object is in the focal plane: An imperceptible real image is formed at infinity. It is
inverted and highly magnified relative to the object.
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 3
(6) Object between the pole and focal plane: A virtual image is formed behind the mirror.
It is erect and magnified relative to the object.
Mirror equation:
Consider a concave mirror
producing a real inverted image
u
M
A‟B‟ of an object AB placed
A
between the principal focus F and
Bꞌ
Principal
the centre of curvature C. The ray
F
P
B
axis
AM parallel to the principal axis is
reflected through the focus. The
Mꞌ
ray AC though the centre of
f
curvature meets this ray at A‟ as
Aꞌ
v
shown in following ray diagram.
Using new sign convention, we have: object distance = PB = - u
Image distance = PB‟ = - v
Radius of curvature = PC = - R
The incident ray AP is reflected along PA‟, governed by the laws of reflection: PC
being the normal at P,
∠ APB (the angle of incidence) = ∠A‟PB‟ (the angle of reflection).
Hence the right-angled ∆s ABP and A‟B‟P are similar.
∴
B′ A′
BA
=
P B′
PB
=
−v
−u
=
v
u
- - - - - - (i)
Also ∠ ACB = ∠ A‟CB‟ (vertically opposite). Hence the right-angled ∆s ABC and
A‟B‟C are similar. . ∴
B′ A′
BA
=
C B′
CB
since all distances are measured from the pole P, we
have: CB‟ = PB‟ - PC = (- v) – (2f) = 2 f – v and CB = PC – PB = (- 2f) – (- u) = u – 2 f.
. ∴
B′ A′
BA
=
R −v
u −R
=
2f −v
u −2f
- - - - - - - (ii)
Equating equations (i) and (ii) we get:
1
1
1
v
u
f
v
u
=
2f −v
u −2f
or 2 f u + 2 f v = 2 u v
∴ + = - - - - - (iii) this is the mirror equation.
Refraction at a single curved surface:
Let APB is a transparent curved surface separating two media of absolute R.I. n1 and n2
such that n2 > n1. Let O is the object placed
on the axis of surface. Let OD is the incident
A
ray which incident obliquely on the surface,
D
i
bend towards the normal after refraction and
r
travels along DI. Another incident ray which
γ
β
α
incident normally on the surface and goes
P
C
O
I
without deviation along PI. These refracted
rays meet at the point I, which is the real
(n1)
(n2)
image of the object O, as shown in figure.
B
Let n be the R.I. of the second medium w.r.t
the first medium.
n
Then, n = 2 .
n1
And according to Snell‟s law, n =
sin i
sin r
⇒
n2
n1
=
sin i
sin r
∴ n1 sin i = n2 sin r ------ (i)
If the point D is very close to the point P, all the angles are very small. Therefore we
write: sin i = i and sin r = r in radians.
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 4
∴ n1 i = n2 r -------(ii)
In ∆ DOC, as i is an exterior angle we write:
i=α+γ
In ∆ DIC, as γ is an exterior angle we write: γ = r + β or r = γ – β. With these values eqn.
(ii) becomes: n1 (α + γ) = n2 (γ – β)⇒ n1 α + n2 β = (n2 – n1) γ ------ (iii)
As angles are very small, they can be expressed in radians in the following manner
assuming that arc PD is approximately straight: α =
With these values eqn. (iii) becomes: n1
n1
OR
PO
+
n2
PI
=
n2− n1
arc PD
+ n2
PO
arc PD
PO
arc PD
PI
,β =
arc PD
PI
= (n2 – n1)
,γ =
arc PD
PC
.
arc PD
PC
------ (iv)
PC
Now, according to new Cartesian sign convention,
Object distance, PO = - u; Image distance, PI = + v and Radius of curvature, PC = +R
n
n
n −n
n1
n
n −n
With these values eqn. (iv) becomes: 1 + 2 = 2 1 ∴
− 2 = 1 2 this
−u
v
R
u
v
R
equation is relation between u, v and R for refraction at single curved surface.
Lens equation / Lens – maker’s equation:
Consider a thin convex lens of the transparent material having R.I. „n‟ is placed in
air medium (n = 1). Let C1 and C2 are
A B
the centers of curvature and R1 and R2
are the radii of curvature of the two
surfaces of the lens.
Let a point object O is placed on the
•
•
I C1 I’
P
axis of lens. The incident ray OP goes O C2
(n)
through lens without any deviation
while the another ray OA will refracted
u
v
along AB by one surface of the lens. In
v’
the absence of the other surface it will
meet the axis at I‟, which would be the
image produced by the first curved surface. Let v‟ be its distance from P.
Thus for the first refraction: n1 = 1, n2 = n, object distance is u, image distance is v‟
and radius of curvature is R1.
∴
1
u
−
n
v′
=
1− n
---- (i)
R1
[∵
n1
u
n2
−
v
=
n1− n2
R
]
But at the point B ray suffers the second refraction from denser medium to rarer
medium along BI, so that I is the final image of the object O. For the second refraction,
I‟ acts as the object. Thus for the refraction at second spherical surface: n1 = n, n2 = 1,
object distance is v‟, image distance is v and radius of curvature is R2.
∴
n
v′
1
n− 1
v
R2
- =
1
v
---- (ii). Combining equations (i) and (ii) we get:
1
1
u
R1
- = (n – 1)
−
1
R2
----- (iii)
Now, if the incident rays are the parallel rays then u = ∞ and v = f, focal length,
∴ eqn. (iii) becomes:
1
f
1
= (n – 1)
R1
−
1
R2
----- (iv). This equation is the lens
equation or the lens maker‟s equation.
1
1
1
f
v
u
*
Form equations (iii) and (iv) we get: = - .
*
As per new sign convention for the convex lens, R1 is +ve and R2 is –ve,
1
1
1
∴ for convex lens
= (n – 1)
+
f
[XI/PHY-II/RSG/ELECTROSTATICS]
R1
R2
Page 5
*
1
For concave lens, R1 is –ve and R2 is +ve, ∴ for concave lens
f
= - (n – 1)
1
R1
+
1
R2
*
Above considerations proves that the focal length of the convex lens positive while that
of the concave lens is negative.
Concept of conjugate foci:
“The conjugate foci are any two points on the axis of a lens, such that if an object is
placed at one point, its image is formed by the lens at the other point and vice versa.”
Magnifying power of a lens (Simple microscope):
D.D.V. – The minimum distance of an object from an unaided eye, at which the object
can be clearly seen without causing strain to the eye is called the distance of distinct
vision (D.D.V.).
“A convex lens with short focal length is used as a simple microscope.” It is also
known as magnifying glass or reading lens. The formation of image by simple
microscope is as shown in figure (a)
The magnifying power of a
simple microscope is defined as, A1
“The ratio of the angle subtended
at the eye by the image to the
A
angle subtended at the eye by the
object, if it is placed at the
β
D.D.V.”
B
B1
P
F
∴ From fig. M.P. = β / α --- (i)
From fig (a), β = A1B1 / PB1
u
D
= AB / PB
(a)
= AB / u where u A
α
α
is the distance of the object from
α
the lens.
B
P
D
From fig (b), α = AB / D
α
α
α
(b)
∴ M.P. = β / α = D / u.
1
1
1
f
v
u
For the convex lens, = -
α
As per new sign convention, u and v both negative while f is positive.
∴
1
f
1
1
v
u
+ =
OR
D
f
+
D
v
=
D
u
⇒M.P. =
D
f
+
D
v
. This is the equation for magnifying power
of the simple microscope which represents that the M.P. of the simple microscope
depends on the image distance v.
*
(i) If the image is formed at infinity then v = ∞. ∴ M.P. =
D
f
+
(ii) If the image is formed at D.D.V. then v = D ∴ M.P. = 1 +
D
∞
D
=
D
f
f
Magnifying power of a compound microscope:
The simplest form of the compound microscope consists of two convex lenses called
objective and the eyepiece. The objective has small focal length and small size. It is
fixed at one end of a hollow metal tube. The eyepiece has a little larger focal length and
size. It is fixed at one end of another hollow tube, which can slide over the other tube.
By a rack and pinion arrangement the distant between the two lenses can be change.
The formation of image by a compound microscope is as shown in figure.
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 6
uO
vO
ue
A
B2
B1
A
O
β
Fe
A1
E
A2
*
Ve
The magnifying power of the compound microscope is defined as, “The ratio
of the angle subtended at the eye by the image to the angle subtended at the eye by the
object, if it is placed at D.D.V.”
∴ M.P. = β / α. But from the figure, β = A1B1 / ue and α = AB / D where ue is the
object distance for the eyepiece and D is the D.D.V.
∴ M.P. = β / α = (A1B1 / AB) x (D / ue)
But A1B1 / AB = vo / uo = Mo = magnification produced by the objective,
also D / ue = Me = Magnification produced by the eyepiece. ∴ M.P. = Mo x Me
If the final image is formed at infinity then:
Me = D / fe where fe is the focal length of eyepiece; Mo = vo / uo
∴ M.P. = Mo x Me = ( vo / uo ) x ( D / fe ) Or it can be proved that, M.P. =
*
fo
D
u o −f o
fe
If the final image is formed at D.D.V. then Me = 1 + D / fe
In this case, M.P. = Mo x Me = (vo / uo) x [1 + (D / fe)]
Or it can be proved that, M.P =
fo
u o −f o
1+
D
fe
Magnifying power of telescope:
The optical device used to see the distant objects clearly is called the telescope.
In the simplest form of the telescope, it consists of two convex lens called objective
and eyepiece. The objective has a large focal length and large diameter, while the
eyepiece has small focal length and small diameter. Each lens is fitted at one end of a
hollow metal
fe
fo
tube. One tube
can slide over
the other. By a
B
α
rack and pinion
α
β
arrangement,
the
distance
A
E
between
the
O
lenses can be
A
changed. The
formation of the image by the telescope is shown in the figure.
“Magnifying power of a telescope is defined as the ratio of the angle subtended at
the eye by the final image to the angle subtended at the eye by the object.”
M.P. =
Angle subtended at the eye by the final image
Angle subtended at the eye by the object
=β/α
From figure, β = AB / fe and α = AB / fo ∴ M.P. = fo / fe ∴ M.P. =
[XI/PHY-II/RSG/ELECTROSTATICS]
Focal length of objective
Fo cal length of eyepiece
Page 7
*
*
When the object is viewed through the compound microscope then it must be brightly
illuminated because----The magnifying power of a compound microscope is directly proportional to the
linear magnification (m0) produced by the objective. For large m0 the distance for the
objective u0 must be small. Hence for a real image, the objective must have a very short
focal length so that the object can be placed just beyond the focal length.
For such a short focal length, the size of the lens and consequently, its aperture or
light gathering power is also small. Hence the object to be viewed must be brightly
illuminated.
Neat labeled schematic ray diagram for a reflecting telescope:
Rays from
distant star
E
M’
T
M
T – Telescope, M – Concave paraboloidal primary mirror,
M‟ – Convex hyperboloidal secondary mirror, E - Eyepiece
*
*
*
*
Can a virtual image be photographed by a camera?
Yes, if we can see a virtual image, we can also photograph it. After all, a camera
works the same way as our eye, just as our eye forms a real image on the retina, a
camera lens forms a real image on the film (or on a charge-coupled device of a modern
camera). Of course the camera has to be in the same position as the eye which sees a
virtual image.
If a thin lens is dipped in water, will its focal length change?
The focal length of a lens depends on the refractive index of the material of the lens
relative to the surroundings as given by the lens maker‟s formula for a thin lens. If a
thin lens is dipped in water, the refractive index 1n2 will decrease and hence its focal
length will increase and the power of the lens will decrease.
Suppose the upper half of a convex lens is painted black. What change is produced in
the real image formed by the lens?
The intensity of light in the image will be halved.
Distinguishing points
Concave mirror
Convex mirror
1. It is a spherical mirror with the inner 1. It is a spherical mirror with the outer
surface reflecting
surface reflecting.
2. It produces either a real and inverted or 2. It produces only a virtual image
virtual and erect image, depending on
regardless of the object distance
the object distance
3. A real image may be diminished of the 3. A virtual image is always diminished
same size or magnified relative to the
relative to the object.
object; virtual image is always
magnified.
4. Its focus is real and in front of the 4. its focus is virtual and behind the
mirror
mirror.
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 8
1.
2.
3.
1.
2.
3.
4.
Simple microscope
It is usually a single convex lens, or a 1.
converging lens system of short focal
length.
The object is placed within the focal 2.
length of the lens.
The maximum magnification is about 3.
10
Compound microscope
In its simplest form, there are two
convex
lenses,
objective
and
eyepiece.
The object is placed just beyond the
focal length of the objective.
Combined magnification by the
objective and the eyepiece is much
larger than that of the simple
microscope. Use of a oil-immersion
type of lens system as objective in a
high quality microscope can yield a
maximum magnification of the order
of 1000.
Compound microscope
The focal length and aperture of the
objective are respectively smaller then
the focal length and aperture of the
eyepiece.
The object is placed just beyond the
focal length of the objective.
It is focused by adjusting the object
distance from the objective.
It is used to observe tiny objects (like
blood corpuscles, animal and plant
cells, and microbes) which, even at
maximum strain-free accommodation,
are invisible to the naked eye.
Telescope
The focal length and aperture of the
objective are respectively far greater
than the focal length and aperture of
the eyepiece.
The object distance is practically
infinite.
It is focused by changing the distance
of the eyepiece from the objective.
It is used to observe huge objects
which appear tiny or faint because of
their huge distances from the eye.
1.
2.
3.
4.
Refracting astronomical telescope
Reflecting astronomical telescope
1. The objective is a converging lens 1. The objective is a front-silvered
system.
concave paraboloidal mirror.
2. Difficulties in making and mounting 2. Flawless mirrors of large apertures are
large lenses limit the aperture to about
easier to make and mount. Hence a
a metre in diameter.
large reflecting telescope can be used
to see fainter, i.e. more distant, stars
and probe deeper into space and time.
3. A lens, especially one with a large 3. A mirror is free from chromatic
aperture, has inherent defects, e.g.
aberration while spherical aberration is
spherical and chromatic aberrations
minimised by using an aspherical
which blurs an image.
(paraboloidal or hyperboloidal) mirror.
* * * * *
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 9
FORMULAE AT A GLANCE
1
1
1
f
v
u
1.
Spherical mirror: f = R / 2; = +
2.
Single spherical refracting surface:
3.
Thin lens: = - = (1n2 – 1)
4.
Simple microscope: M = D / u = D / f (image at ∞) = 1 + D / f (image at the least
1
1
1
1
f
v
u
R1
n2
−
v
-
n1
u
1
=
n2− n1
R
,m=
R2
hi
ho
v
1
1
1
u
f
f
f1
= ;p= ; =
+
1
f2
+----
distance of distinct vision)
5.
=
=
6.
v0
Compound microscope: M = m0 x Me =
f0
D
f0+ u0
fe
v0
uo
Astronomical telescope: M =
uo
f0
fe
x 1+
D
fe
x
D
ue
=
v0
uo
x
D
fe
(image at infinity)
=
f0
f0+ u 0
1+
D
fe
(image at D)
, L = f0 + fe (normal adjustment)
* * * * *
[XI/PHY-II/RSG/ELECTROSTATICS]
Page 10
SOLVED PROBLEMS
1. A convex lens has focal length 20 cm, which produces an image four times larger than
object. Calculate the possible positions of objects.
Solution: MP = v/u = Image size / Object size = 4. ∴ v = 4u
Let object distance is x, then image distance = 4x.
1
1
1
1
v
u
0.2
= - ⇒
f
1
=
4x
-
1
1
=
−x
1
5
x
4x
+ =
4x
⇒ x = 25 cm. As this is object distance and for the
real image it is to the left of lens it is negative ∴ u = - 25 cm. While for the virtual image,
u = - x, v = - 4x
1
f
1
1
1
v
u
0.2
= - ⇒
1
=
− 4x
-
1
1
1
x
4x
= -
−x
=
3
4x
⇒ x = 15 cm. ∴ u = - 15 cm.
2. A Plano - convex lens of glass has radius of curvature of 30 cm. If the glass has
refractive index of 1.6, find the focal length of lens.
1
1
f
R1
Solution: = (n – 1)
−
1
R2
= 0.6 x
1
30
⇒ f = 50 cm
3. A convex lens used as simple microscope has focal length of 2.5 cm. Find its magnifying
power for the image at DDV and the position of the object.
Soln: M = 1 +
D
= 1 + (25/2.5) = 11 Also, M = D / u ⇒ u = D / M = 2.27 cm. As this is
f
object distance it is negative. ∴ u = -2.27 cm.
4. A compound microscope has an objective and eyepiece of focal lengths 2.5 cm and 6.0
cm respectively. The lenses are 20 cm apart. A small object is kept at 3 cm from the
objective. Find the distance of final image from eyepiece and the magnifying power of
microscope.
Solution: f0 = 2.5 cm; fe = 6.0 cm; L = 20 cm; u0 = - 3 cm; ve = ? and M = ?
1
v0
-
1
u0
=
1
f0
;
1
v0
1
1
3
2.5
+ =
;
1
v0
1
=
2.5
1
- ⇒ v0 = 15 cm
3
∴ ue = L – v0 = 20 – 15 = 5 cm. Now,
MP = M0 x Me =
v0
u0
x
ve
ue
=
15
3
x
30
5
1
ve
=
1
ue
-
1
fe
=
1
−5
1
1
6
30
- =
⇒ ve = 30 cm
= 30
5. The objective and eyepiece of an astronomical telescope are 60 cm apart. If the
magnifying power of telescope is 19, find the focal lengths of both the lenses.
Solution: M = f0 / fe ⇒ f0 = 19 fe. And for telescope L = f0 + fe
Solving these two equation we get: f0 = 57 cm and fe = 3 cm.
6. A compound microscope has a magnifying power of 40. Assume that the final image is
formed at DDV. If the focal length of eyepiece is 10 cm, calculate the magnification
produced by objective.
Solution: MP =
v0
u0
1+
D
fe
⇒
v0
u0
= MP / 1 +
D
fe
= 40 / (1 + 2.5) = 11.44
7. A convex lens has focal length of 2.0 cm. Find the magnifying power if image is formed
at DDV.
Solution: MP = 1 +
D
fe
= 1+
[XI/PHY-II/RSG/ELECTROSTATICS]
25
2
= 13.5
Page 11
8. A glass slab has concave surface of radius of curvature 2.0 cm. The glass has refractive
index of 1.5. If a point object is placed on axis in air at a distance of 18 cm from concave
face, find the position and nature of the image.
Solution:
μ2
v
-
μ1
=
u
μ2 − μ1
R
∴
1.5
v
-
1
− 18
=
0.5
−2
1
= - ⇒ v = - 4.907 ≈ - 4.9 cm.
4
As the (-ve) sign, the image is virtual.
9. An objective is placed at 15 cm from a convex mirror having the radius of curvature 20
cm. Find the position and kind of image formed by it.
Solution: f = R / 2 = 20 / 2 = 10 cm.
1
f
1
1
1
1
1
1
v
u
𝑣
𝑓
u
10
= + ⇒ = - =
-
1
− 15
= 1 / 6 ∴ v = 6 cm
The image is formed at a distance of 6 cm from the mirror; the positive
value indicates that it is virtual, erect and behind the mirror.
10. An optical system uses two thin convex lenses in contact having the effective focal
length of 30/4 cm. If one lens has focal length of 30 cm, find the focal length of the other
lens.
1
1
𝑓
f1
Solution: =
+
1
f2
∴
1
f1
=
1
𝑓
-
1
f1
=
1
30 4
-
1
30
=
1
10
∴ f1 = 10 cm
* * * * *
[XI/PHY-II/RSG/ELECTROSTATICS]
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