Download Eye – pieces (Oculars) and their Cardinal Points Paper: Optics

Eye – pieces (Oculars) and their Cardinal Points
Paper: Optics
Lesson: Eye – pieces (Oculars) and their Cardinal Points
Author: Dr. D. V. Chopra
College/Department: Associate Professor (Retired),
Department of Physics and Electronics, Rajdhani College,
University of Delhi
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Eye – pieces (Oculars) and their Cardinal Points
Objectives: After studying this chapter you should:
1.
2.
3.
4.
5.
Be able to know various types of eyepieces such as Ramsden’s eyepiece,
Huygens’ eyepiece and Gauss’ eyepiece.
Be able to understand functions of eye lens and field lens
Be able to understand optical properties of and to study the cardinal points of
Ramsden’s eyepiece and Huygens’ eyepiece
Be able to know the construction and use of Gauss’ eyepiece in spectrometer
telescope
Be able to work problems involving eyepieces and their cardinal points
1. Eye – piece:
The objective of a telescope as well as a microscope is a convex achromatic doublet
which produces a real and inverted image of the object. Such an image is diminished in
size in the case of a telescope and magnified in the case of a microscope. The function of
an eye – piece in the optical instrument is to magnify the real image formed by the
objective .
An eye – piece is a specially – designed co-axial system of two lenses, separated by
a distance. It is also called compound eyepiece.
(a) Constructions: An eye – piece consists of two convex lenses placed co-axially with the
axis of the instrument. The lens facing the objective is called field lens and the other
lens placed near the eye is called eye-lens. Such a combination constitutes an eye-piece
(also called an ocular).
(b)Difference between an ey-piece and a magnifier: A magnifier is a convex lens or a
lens system used to magnify a real object. An eye – piece is used to magnify a real
image (produced by an optical system) formed before it.
(c) An eye – piece consisting of a single converging lens suffers from the following
defects:
(i). In order to increase the magnification, the single lens of the eye-piece must have
small focal length and it will have fairly curved surfaces. Such a magnified image
suffers from defects such as chromatic aberration, spherical aberration, etc.
(ii). The field of view is too small to see the outer parts of the image. Some of the rays
from the image formed by the objective are not able to reach eye through the single
lens of the eye-piece. This is shown in Fig.1, A’B’ is the real image formed by the
objective and E.L. is the single-lens eye-piece. Rays with the single arrow from A’B’
are not entering the eye which is placed closed to the eye.
Both these defects can be reduced by using two lenses in an eye-piece. Such a
combination of two convex lenses is called an eye-piece (Fig.2). The lens nearer to the eye
is called eye-lens (E.L) and the other which faces the real image formed by the instrument
is called field lens(F.L): the combination of the two is so designed as to reduce the defects
of images such as spherical aberration, astigmatism, distortion and chromatic aberration,
this latte being by far the most important in the case of a single lens
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Eye – pieces (Oculars) and their Cardinal Points
Position of
eye
E.L
Fig.1 A single-lens eyepiece
Position of
eye
The image
formed by the
optical instrument lies infront of field
lens (F.L) and all the rays from it are
able to enter the eye, placed close to
E.L
the eye – lens(E.L.) . The eye – lens
F.L
is placed at the same position as an
eye – piece consisting of single lens
Fig. 2 A Compound Eye-piece
would do, but the position of the
field lens is either in front of, or
behind, the image
formed by the
objective. But, it is required that the final image formed by the eye – piece should be at
infinity so that it may be seen without any strain to the eye. For this requirement, the
image formed by the field lens must lie in the focal plane of the eye – lens so that the rays
are rendered parallel on passing through E.L.
Function of eye lens (E.L.):
Its purpose is mainly to magnify the image due to the field lens and to form it at
infinity for normal vision. It is called eye lens because it is placed close to the eye.
Functions of field lens (F.L.):
(i). In the combination with the eye – lens, it decreases sufficiently the focal length of the
eye – piece. This way, it increases the magnification of the final image seen by the eye.
(ii). It increases the field of view because it bends the rays form the other parts of the
image
towards its axis and these rays enter the eye after passing through the eye
lens. Hence, rays from outer parts of
are not lost and the whole image
is seen.
It is called field lens because it increases the field of view.
(iii). In combination with the eye lens, the eyepiece can be made wholly or appreciably free
spherical and chromatic aberration.
(iv). Eye can be placed at the centre of the eye – lens, thereby increasing the angular
image field and thus whole of the image is seen.
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Eye – pieces (Oculars) and their Cardinal Points
As stated above, as the final image through eyepiece is to be seen at infinity, the
image
as well as the field lens (Fig. 2) must coincide with the first focal plane of the
eyelens. This is, theoretically, the best location for the field lens. This is so because in this
case u = 0, and v = 0, and thus there is no change in the magnification and also, in
addition, it will increase the field of view. The disadvantage of this location is that any
scratches or dust particles on the surface of field lens or its other blemishes (like small air
bubbles inside it, etc.) are observed magnified along with image
and thus reduce the
clarity of the image. In practice, the field lens is placed either a little in front of image
(as in Huygens’ eye – piece) or a little behind it (as in Ramsden’s eye – piece).
Position of a cross – wire:
In order to make some measurements on the final image formed by the eyepiece,
we require a cross-wire or a micro meter scale to be placed inside the eye piece. A crosswire must fulfill the following conditions:
(i). It must be placed coincident and in focus with the final image.
(ii). It must be magnified by both the lenses of eye – piece.
As the final image formed by the eye – piece is at infinity, therefore it must lie in the
first focal plane of the eye – piece. Therefore, the image formed by an objective of any
optical instrument like telescope and microscope must lie in the first focal plane of the eye –
piece.
If the first focal plane of the eye – piece lies outside the eye – piece in the object
space i.e. in front of the field lens, then real object (i.e. cross-wire) can occupy the position
coincident with the final image formed by the objective of the instrument. Such an eye –
piece is called positive eye – piece since this eye – piece requires a real object (i.e. crosswire) to be presented to it, (i.e. since the first focal plane of the eyepiece is on the positive
side of the field lens).
If the first focal plane of the eyepiece lies within the two lenses of the eye-piece,
than a virtual object or cross-wire can occupy the position coincident with the final image
formed by the objective. Such an eye piece is called negative eye – piece as the first focal
plane of the eye piece is on the negative side of the field lens.
Since this eye – piece
requires a virtual object (i.e. virtual cross-wire) to be presented to it, such an eye – piece
cannot be used with corss-wires as the positive eye – piece can.
The eye – pieces which are commonly employed are as follows:
(i). Ramsden’s eyepiece
(ii). Huygens’ eyepiece
(iii). Gauss’ eyepiece
These compound eyepieces have been discussed in the next sections.
2. Ramsden’s eyepiece:
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Eye – pieces (Oculars) and their Cardinal Points
Construction (Fig. 3):
It is consists of two
thin plano – convex lenses,
F.L. and E.L., of some
material and focal lengths, f,
placed
separated
by
a
distance
, as shown in
F1
L
F.L
E.L
f
f
eye
x
axis
L1
L2
y
z
Fig. 3
Position of
eye
They
are
placed
coaxially with their convex
surfaces facing towards each
other. The lens E.L. placed
close to the lens is eye – lens
Fig. 3. Ramsden’s eyepiece
while the lens F.L. placed
away from the eye is fieldlens.
is the image formed by the objective of the optical instrument. It coincides with
the first focal plane F of the eye – piece, as shown in Fig. 3. Since F1 is on the positive side
of field lens F.L. , this eye – piece is called positive eye – piece. The F1 or
is also the
position of cross – wires. Only narrow beam passes through the eye – piece. The various
aberrations are reduced to tolerable amounts in the following way.
Condition of achromatism: Chromatic aberration has to be eliminated in the sense that
the size of the different coloured images the same.
Condition of achromatism for two lenses of focal lengths f1 and f2 is given by the distances
of separation,
In the present case, f1 = f2 = f
That is to say, the field lens F.L. should be placed in the focal plane of the eye – lens
E.L. Because eye – lens is focused on the field lens and so any scratches, or defects, on its
surface are seen and spoil the look of the final image. Therefore, instead of distance x = f,
the distance
between the two lenses of the eyepiece is kept. It causes slight
departure from achromatism.
chromatic aberration.
Hence, Ramsden’s eyepiece is not completely free from
Spherical aberration is reduced by sharing the deviations as equally as possible
among four refractions of the two lenses, F.L. and E.L. This helps to reduce coma as well.
Spherical aberration is further reduced to minimum by taking each individual lens planoconvex (F.L. and E.L.). The radius of the circle of least confusion due to spherical
aberration and astigmatism is reduced by the fact that only narrow beam from the objective
is allowed to pass through the eyepiece. Curvature of the field and distortion are also small
over fields of 400.
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Eye – pieces (Oculars) and their Cardinal Points
Working :
The
, formed by the objective, will serve as real object to the eyepiece. For normal
vision, this image
should be situated at the first focal plane F 1 of the eye – piece so that
the final image formed by the eye – piece, may lie at infinity. Actually, however, the final
image at infinity is formed in two steps.
In the first step, the real image
formed by the objective serves as a real object
to the field lens (F.L.) which forms its virtual image at
which is also the first focal plane
of the eye – lens (E.L.). In the second step,
serves as an object to the eye – lens
(E.L.) as
is at the first focal plane of the eye – lens, the image will be formed at
infinity.
Optical Properties of Ramsden’s – eyepiece:
By optical properties of any optical system we mean those properties or
characteristics of it on which its optical behavior depends e.g., the focal length, power,
positions of cardinal points, angular field of view, aberration, etc. The various aberrations
have already been described in this section. Other optical properties have been described
below.
Focal length of the eye – lens = f =L2
Distance
Focal length F of the equivalent lens is given by
Substituting f1 = f2=f and x = 2f/3
Position of equivalent lens is shown by L with the Fig.3 Since field lens forms image
, focal length f of field lens is given by
where u = L1
of
and v= L1
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Eye – pieces (Oculars) and their Cardinal Points
This gives
This indicates that the real image
formed by the objective lies at a distance
in front of
the field lens:
Position of cross-wires (or micrometer scale):
The image of the object formed by the objective lies in front of the field lens, in the
first focal plane of the eye lens, which is real, and cross-wires can, therefore, be used. For
this reason it is called a positive eyepiece.
Position of cross-wires coincides with
at a distance
in front of field lens of focal
length f. This eye – piece magnifies both image
and cross-wires equally when seen
through the eye – lens. Hence accurate measurements can be taken.
Cardinal Points of Ramsden’s eye – piece (Fig.4)
Position of first principal Point (P1)
Distance of the first principal point P1 from field lens L1 is given by Eq. 3.9-8. It is
where x = distance between the
two lenses L1 and L2 .
=
Equivalent
lens
Incident
light
F.L(f)
axis
L1
F1
L
E.L(H)
L2
N2 P2 N1 P1
F2
F = focal length of equivalent lens
(from Eq. 2-2)
f2 = focal length of the eye – lens L2
Fig.4.
Cardinal points of
Ramsden’s eyepiece
=f
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Eye – pieces (Oculars) and their Cardinal Points
P1 lies to the right of L1, as shown in Fig.4
Position of second principal point (P2):
Using Eq. derived in chapter on cardinal points of a coaxial system, L2P2 =  =
This P2 lies to the left of L2 ; as shown in Fig 4. The two principal planes through P 1
and P2 are real (i.e. lie within the system), but crossed.
Position of focal points (F1 and F2)
Distance of the focal point, F1 from the field lens L1 is given by Eq. (3.9-11)
Thus, F1 lies to the left of the field lens L1 at a distance
from it f2 = f is the focal length of
eye – lens. Distance of second focal point F2 from the eye – lens L2 is (see chapter on
Cardinal Points of a coaxial system)
Thus, F1 lies to the right of the eye lens L 2 at a distance
from it where f1 = is the focal
length of the field – lens L2 , as shown in Fig 4. Since the first focal point (and hence the
first focal plane) is real here, the eye – piece can be used to magnify a real object, i.e. it
can be used as a simple microscope or magnifier.
Position of nodal points:
The medium on either side of the eye – piece is same (i.e. air). Hence, nodal points N1 and
N2 coincide with principal points P1 and P2 respectively.
Dotted vertical lines through (F1, F2), (P1, P2) and (N1, N2) s how respective planes in
Fig 4. These planes are normal to the axis of the eye – piece.
Position of Equivalent lens (L):
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Eye – pieces (Oculars) and their Cardinal Points
The distance of the equivalent lens (L) behind the first lens L1
Thus the equivalent lens (L) should be placed coinciding with the first principal plane
(P1). From Eq(2-2),
Focal length of the equivalent lens,
Power of Ramsden’s eye – piece:
where f is measured in metres. The power is positive. Since the eyepiece has the same
medium on either side of it [i.e. on left of L1 and right of L2], the first and second focal
lengths are both equal to each other, so that each is numerically equal to
, with the first
focal length negative and the second, positive. From Fig. 4, we have
Angular field of view is about 300, but the correct field is smaller than in Huygens’ eye –
piece.
When a Ramsden eyepiece is used in an astronomical telescope with an objective of focal
length F0, the magnifying power of the telescope is given by
It is to be noted that the magnifying power of a Ramsden eyepiece when
used as a magnifier
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Eye – pieces (Oculars) and their Cardinal Points
Importance of Ramsden’s eye – piece:
It can examine a real object or a real image. Therefore, it can be fitted with a crosswire or a micrometer scale. It is used in optical instruments where accurate measurements
are required. It is used in spectrometry to analyse each individual colour the spectrum,
using white light.
Although the eyepiece is not free from chromatic aberration; in
spectrometry this aberration does not effect the measurements.
Best position for the eye to be placed:
From Fig 3, it can be easily gathered that the rays from the upper extremity of the
objective pass through Z after emergence from the eye – piece, those proceeding from its
lower extremity pass through X and these through its centre pass through Y. Thus, XYZ is
obtained as the image of the objective, as formed by the eyepiece, (shown in Fig 3) and all
rays travelling ethrough the
objective pass through XYZ. It is
called the eye – ring (or the exit
about
pupil), with its centre Y, called
6mm
the eye – point. All rays which
Cross-wire
enter through the objective must
E.L
F.L
Eye-ring
emerge through the image of the
objective formed by the eyepiece
if they emerge at all. This image
Incident
O
Eye-pupil
is called the eye – ring or exit
light
F1
pupil. If the eye be kept at the
eye-ring, the maximum number
Ebonite cap
of rays emerging from the
B
objective pass through the eye.
In
the
case
of
normal
magnification, the size of the
Fig. Ramsden’s eyepiece with cross section
eye-ring is the same as that of
the eye-pupil, so that all rays
travelling from the objective pass
through the eye and the object appears the brightest. This is the best position of the eye,
because this is the place where the rays on leaving the eyepiece come closet together. The
aperture of the objective is called the entrance pupil.
The position of the eye ring is usually marked by a small circular opening in the
ebonite cap of the brass tube carrying the eyepiece. Such an eyepiece is shown in Fig. O
is the small circular opening in the ebonite cap of the brass tube B which carries the field
lens (F.L.), eye-lens (E.L.) and cross-wires, F1, infront of field lens (F.L.). The distance
between the eye-ring and the eye lens is called the eye clearance. This must be at least
6mm to allow free movement of the eye-lenses. For persons using spectacles , it should
obviously be a little more. As the cross-wire are situated outside the eyepiece, infront of the
field lens, the distance between the eyepiece and cross-wires can be changed and adjusted
for person of different vision. Hence, persons of different vision can use the same eyepiece.
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Eye – pieces (Oculars) and their Cardinal Points
3. Huygens’ eyepiece:
The Hygens’ eyepiece was introduced in order to improve the corrections for both
spherical and chromatic aberration of the Ramsden’s eyepiece. The astronomical telescope
is fitted with a Huygens’ eyepiece.
Construction:
This eyepiece consists of two thin plano-convex lenses L1 and L2 of the same
material but of different focal lengths. The lens L1 is field lens (F.L.) of focal length 3f
whereas the lens L2 is eyelens (E.L) of focal length f.
L
F.L(3f)
F1
E.L(f)
eye
axis
L2
A A
L1
B
B
f
2f
3f
Fig.6. Huygens’s eyepiece
These lenses are kept coaxially separated by a distance x = 2f, with their convex surfaces
facing the incident rays, as shown in Fig.6. As the first focal plane F 1 of the eyepiece is on
the negative side of the field lens L1, this eyepiece is called negative eyepiece. Only narrow
beam of light passes through the eyepiece. This eyepiece satisfies the conditions for both
minimum spherical aberration and (lateral) achromatism. It is, therefore, known as a
theoretically perfect eyepiece.
Working :
The image
is the image formed by the objective of the optical instrument. It
serves as a virtual object to the eyepiece. For normal vision, this image
is to be formed
at the first focal plane F1 of the eyepiece, so that the final image formed by the eyepiece
may be at infinity. Actually, the formation of final image at infinity takes place in two steps.
In the first step, the converging rays from the objective, before they converge at
, are
further refracted by the field lens (F.L.) and a real image is formed at
and not at
.
The focal plane of the eyelens (E.L.) coincides with
. Thus, for field lens,
acts as
its virtual object while
is its real image. In the second step,
serves as an object
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Eye – pieces (Oculars) and their Cardinal Points
to the eyelens (E.L.). As
is formed at infinity.
lies at first focal plane of the eyelens (E.L.), the final image
Let us apply the lens formula
to the field lens of focal length 3f.
The image
distance
is situated at a distance
to the right of the field lens and hence a
, to the left of the eyelens.
Hence, it must lie in between the two
lenses at a distance equal to half of its focal length from either lens. A field stop is usually
placed in the position of
to make the final image more sharply defined.
Optical properties of Huygens’ eye piece:
We shall describe below the focal length, power, aberration, cardinal points, angular
field of view, etc.
Focal length F of the equivalent lens is given by
Substituting f1 = 3f, f2 = f, and x = 2f, we have
Position of equivalent lens is shown by L. in Fig. It is situated at a distance
the first focal plane F1. Now, distance,
coincides with
from
.
It is so because the focal plane of L2
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Hence
Since
Eye – pieces (Oculars) and their Cardinal Points
For distinct vision, the image
due to the objective must be formed within the first
focal length LF1 of the eyepiece (i.e. the image
due to the field lens must be within the
first focal length of the eyelens) so that the final image may be formed at the least distance
of distinct vision.
Position of cross-wires (or Position of micrometer scale):
If a cross-wire or a micrometer scale is to be fitted, it should be placed coinciding
with
where the real image is produced after successive refractions through the
objective (not shown in Fig.5) and the field lens (L 1). Thus, the cross-wires will be
magnified by eyelens (L2) only while the image
is formed by both the lens L1 and L2 of
the eyepiece.
The magnification of the image and cross-wires will nto be same.
Consequently, the image of the cross-wires will not be of a very much inferior quality, due
to aberrations and distortions. Thus, cross-wires at
cannot be used with this eyepiece
to make measurements. In fact, since this eyepiece requires a virtual object
to be
presented to it, it cannot be used with cross-wires as the Ramsden eyepiece. It is often
called a negative eyepiece to distinguish it from the Ramsden’s eyepiece, which is called a
positive eyepiece.
Removal of aberrations:
(i). Condition of achromatism:
It is given by,
In Huygens’ eyepiece, f1 = 3f, f2 = f
This is in complete agreement with the distance of separation, L1L2 = 2f, as shown in
Fig.6. Hence, this eyepiece is completely free from chromatic aberration.
(ii). Condition of minimum spherical aberration:
The condition of minimum spherical aberration is,
In the case of Huygens’ eyepiece,
f1 = 3f and
f2 = f,
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Eye – pieces (Oculars) and their Cardinal Points
This distance of separation is the same as L1L2 = 2f, is shown in Fig. Hence, the Huygens’
eyepiece is free from spherical aberration. It follows from Eq(3-2) and (3-3) that spherical
and chromatic aberration are simultaneously removed.
(iii). It is mentioned above that the connection for chromatic aberration is as good as
can be achieved, and the correction for spherical aberration and coma is better in this
eyepiece than in the Ramsden’s eyepiece. It is also free from astigmatism and distortion to
the same extent, but it suffers from considerable curvature of the field, the curvature being
convex towards the observer’s eye.
Cardinal Points of Huygens’ eyepiece:
Position of principal points (P1 and P2)
Distance of the first principal point P1 from the field lens is given by (See chapter on
Cardinal Points of a coaxial system)
F.L(3f)
L
E.L. (f)
Incident Light
P2
N2
Axis
L1
f
F1
L2
N1 P1
F2
f
f
Fig.7.Cardinal points
Huygens’ eyepiece
2f
3f
Here x = distance of separation betwee the two lenses L1 and L2 = 2f
F = focal length of equivalent lens =
(from Eq.3-1)
f2 =focal length of the eyelens L2 = f
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Eye – pieces (Oculars) and their Cardinal Points
Hence, P1 lies on the right of L1 at a distance 3f. Position of the second principal point P 2
from the second lens L2 is given by Eq. L2P =  = xF/f1 where f1 is the focal length of the
first lens L1; (f1 = 3f)
Hence, P2 lies on the left of L2 at a distance f. The principal points P1 and P2 have been
marked in Fig. 7. The dotted (vertical) lines through P 1 and P2 show respective principal
planes.
Positions of focal points (F1 and F2)
Distance of the first focal point F1 from the field lens L1 is given by Eq.(3.9-11)
Thus, F1 lies on the right of L1 at a distance
, as shown in Fig. 7. Distance of the second
focal point F2 from the eyelens L2 is given (See chapter on Cardinal Points of a coaxial
system).
Thus, F2 lies on the right of L2 at a distance .
Positions of nodal points:
Since the coaxial system of lenses L1 and L2 of the eyepiece has the same medium (air) on
either side of it, the nodal points N1 and N2 coincide with the principal points P1 and P2
respectively.
Dotted vertical lines through (F1, F2) , (P1, P2) and (N1, N2) show respective planes in Fig. 7.
Postitions of Equivalent Lens(L):
Distance of equivalent lens L behind the field lens L1
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Eye – pieces (Oculars) and their Cardinal Points
Thus the equivalent lens L as shown in Fig. 7 should be placed coinciding with the first
principal plane P1. It must be placed at a distance f behind the eyelens. From Eq. 3-1,
focal length of equivalent lens is
Power of Huygens’ eyepiece is
where f is measured in meters.
The angular field of view is rather small about 200 but the corrected field is greater than
Ramsden’s eye – piece.
Discussion: (i) The first and second principal planes (P 1 and P2) lie at distances 3f to the
right of L1 and f to the left of L2 respectively (i.e. at distances f to the right and to the left of
the eyelens respectively), as shown in Fig.7. The two principal planes are thus crossed,
with the first principal plane, virtual (P1) since it lies outside the system.
(ii). The Huygens’ eyepiece cannot be used to see any real object. This is due to the fact
that the first focal plane (and hence the first focal plane F 1 shown in Fig. 6) is virtual, lying
in between the two lenses. This eyepiece cannot be used to magnify a real object but only
an image. It cannot therefore, be used as a simple microscope or a magnifier. Since it can
be used only to examine an image (and not a real object), it is called a negative eyepiece.
Ramsden’s eyepiece alone is used as a magnifier, its focal plane lies outside the
eyepiece. Huygens’ eyepiece is often used in microscopes and in observational instruments
using white light whereas Ramsden’s eyepiece is used in telescopes or optical instruments
with which measurements of distances and angles are to be made using cross-wires or
micrometer scale.
4. Different Eyepiece used in different optical instruments:
(i) Huygens’ Eyepiece in low power microscope:
In low power microscope it is of utmost importance that the image should be
completely free from the defects of chromatic and spherical aberrations. Hence Huygens’
eyepiece should be used in a low power microscope.
(ii). Ramsden’s Eyepiece in the spectrometer telescopes:
In the spectrometer, measurements are to be made and hence chromatic and
spherical aberrations are of secondary importance. Hence, Ramsden’s eyepiece should be
used in the spectrometer telescopes.
(iii) Huygens’ eyepiece cannot be used in Fresnel’s biprism experiment.
A microscope scale for measurement purposes cannot be employed in Huygens’
eyepiece, because the image of the object is formed by both the lenses (field lens and
eyelens) of the eyepiece while that of the scale is formed only by the eyelens. This leads to
unequal magnifications of the image and the scale.
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Eye – pieces (Oculars) and their Cardinal Points
(iv) Huygens’ eyepiece cannot be used to see our palm. It is because this eyepiece cannot
be used as magnifier. The first focal plane of the Huygens’ eyepiece falls within the
eyepiece, where no real object can be placed. This
eyepiece can be used only for viewing the images
formed by objectives of microscope and telescope.
Silicate
Flint
Ramsden’s eyepiece can be used for seeing the
images as it can be used as a magnifier and its focal
plane is situated outside the eyepiece.
(v) Achromatic Ramsden’s eye-piece:
Such an eyepiece is often used in prismatic
binoculars.
The lateral chromatic aberration of a Ramsden’s
eyepiece is slightly greater than that of Huygens’
eyepiece, but its longitudinal chromatic aberration is
about half as great as that of Huygens’ eyepiece.
Kellner used lenses as shown in Fig. 8. Such a
combination improved the chromatic correction.
field
lens
Eye lens
Barium silicate
crown
Fig.8.Acrometric Ramsden
eyepiece
(vi) Compensating oculars:
A compensating ocular is an eyepiece used to correct residual errors of the objectives of
telescopes and microscopes. It is a combination of low power Huygens’ eyepiece and high
power Ramsden’s eyepiece.
5. Relative merits and demerits of Ramsden’s and Huygens’ eyepieces:
Huygens’ eyepiece
1. It is completely free from spherical aberration
Ramsden’s eyepiece
1. It is not completely free from
spherical aberration.
2. It is compleely free from
aberration for all colours
2. It is not completely free from
chromatic aberration.
lateral
chromatic
3. It can be used with white light
3. It can be made achromatic for only
two chosen colours. It cannot be
used with white light.
4. The correction for coma is better
4. The correction for coma is less
5. It is free from astigmatism to the same extent
5. It is free from astigmatism to the
same extent
6. Curvature of the field and distortion are not
small. It suffers from considerable curvature of
the field, the curvature being convex towards the
observer’s eye.
7. It cannot be used with cross-wires since it
6. Curvature
of
the
field
and
distortion are small over fields of
400. Final image is almost flat.
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7. It can be used with cross-wires
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Eye – pieces (Oculars) and their Cardinal Points
requires a virtual object to be presented to it. It
is often called a negative eyepiece. It is,
therefore suitable for only visual or qualitative
purposes.
It is used with microscopes and
telescopes working with white light. It is not
suitable when measurements are required. It is
used in microscopes required for biological work
(no measurements)
8. It cannot be used as a magnifier or a simple
microscope. This is so because the first focal
plane is virtual and lies in between the two
lenses.
since it requires a real object to be
presented to it. It is called positive
eyepiece. It is, therefore, suitable
for quantitative measurements. It
can be used with microscopes,
telescopes
and
spectrometers
working with monochromatic light.
8. It can be used as a magnifier or a
simple microscope.
This is so
because the first focal plane is real
and lies to the left of the field lens.
9. The two principal planes are crossed but the first
principal plane is virtual
9. The two principal
crossed and real.
10. The distance between the eye-ring and eye-lens
is too small for comfortable use of the instrument
to which the eye-piece is fitted.
10. The distance between the eye-ring
and eye-lens is greater; this makes
it more comfortable for the eye to
use.
11. It
has
comparatively
less
longitudinal Chromatic aberration
11. There is considerable amount of longitudinal
chromatic aberration
planes
Value Addition
Christian Huygens (1629-1695)
Christian Huygens was born of a well-to-do and influenced family at the Hague in Holland.
He published a book on optics entitled “Traite dala Lumiere” in 1690. He brought about
great improvements in telescopes and microscopes by removing defects in images formed
by lenses. According to him, the motion of light is in transverse waves. He thoroughly
studied the phenomenon of reflection, refraction and rectilinear propagation of light and
came to the conclusion that light consists of wave motion travelling in all pervading but
hypothetical medium, called ether. However, Hooke believed that light travels in longitudinal
waves but Newton discarded this view. It took more than a century for Young (1773-1829)
and Fresnel (1788-1827) to show the wave nature of light. We now know that the
phenomenon of interference, diffraction and polarization of light can be admirably explained
on Huygens’ wave theory of light.
6. Gauss’ Eyepiece:
It is often used in spectrometer
telescopes. It is useful in focusing the
collimator
and
telescope
of
spectrometer for infinity by Schuster’s
method. It is also helpful in adjusting
the axis of telescope perpendicular to
the axis of the instrument.
S
Q
C
450
L2
G
It is just a Ramsden’s eyepiece with a
side opening between the two lenses of
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F.L.
T
L1
E.L
Fig.9. Gauss’ Eyepiece
Page 18
are
Eye – pieces (Oculars) and their Cardinal Points
the eyepiece and is used in spectrometric measurements. It is shown in Fig.9. It consists
of two lenses L1 and L2, each having equal focal length f and separated by a distance
.
However, the separation between the two lenses should have been
in order to
satisfy the condition of minimum chromatic aberration. For this separation, d = f, the field
lens will be situated at the first focal plane of the eyelens. The objective lens forms the
image of this focal plane. When seen through the eyelens, any scratch or defect of the lens
(F.L.) surface is also magnified along with the final image. This difficulty is overcome by
taking distance
of separation instead of f.
This will create small chromatic
aberration. Such an aberration is further minimized by choosing achromatic doublets.
A thin plate G of unsilvered glass is placed between the two lenses F.L. and E.L. inclined at
an angle of 450 to the axis of the lenses which are placed in a tube T. The tube T has a
small side opening to admit light form the source S of light. Light is reflected along the axis
of the tube by G. Cross-wires (if used) are fixed at C. Sometimes this Gauss eyepiece is
fitted in telescope of a spectrometer. For spectrometer fitted with a Gauss eye – piece the
following adjustment is made:
Light from a source of light (e.g. a candle or a torch) is placed opposite the opening Q.
Light is reflected along the axis of the telescope by glass plate G. These rays illuminate the
cross-wires and proceed further till they emerge out of the objective of the telescope. If
these rays are reflected back into the telescope by a plane reflecting surface, we shall see
two images of cross – wires, namely, one a reflected image and second a direct image when
viewed into the eyepiece. The two images so obtained are used to focus the telescope for
infinity.
7 Solved Examples:
Example 1:
1. An eyepiece is made of two thin lenses, each of 20 mm focal length, separated by a
distance of 16 mm. Find:
(a). the equivalent focal length of the eyepiece
(b). the distance of the focal plane form the nearer lens, and
(c). the angular magnification provided,.
Solution:
(a).
(b).
This image is formed by first lens.
distance of separation is 16mm.
It is formed 4mm beyond second lens because
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Eye – pieces (Oculars) and their Cardinal Points
(c). Angular magnification or magnifying power of an eyepiece is
Example 2:
(a) Two thin plano-convex lenses of same material in an Huygens’ eyepiece are 10 cm
apart. Find the focal lengths of the lenses and the equivalent focal length of the
eyepiece.
(b) Calculate these quantities for a Ramsden’s eyepiece.
Solution:
(a) Huygens’ eyepiece consists of two planconvex lenses, namely, a field lens of focal
length 3f and an eyelens of focal length f, and they are separated by a distance x =
2f. Its equivalent focal length is F.
Because x = 10 cm
 Focal length of field lens = 3f = 3 x 5 = 15 cm
Focal length of eyelens = f = 5 cm
Equivalent focal length,
(b).
For Ramsden’s eyepiece,
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Eye – pieces (Oculars) and their Cardinal Points
Example 3:
An Huygens’ eyepiece has an eye-lens of 4 cm focal length. Sunlight is falling over
the eyepiece. Find the position of the image in this condition and the distance of second
principal point from the eye – lens.
Solution:
In Huygens’ eyepiece,
Given
Equivalent focal length,
Given object is sun which is situated at infinite distance. Its image is formed at the
second focal point F2 of the eyepiece.
Distance of second focal point F2 from the eyelens L2 is
Assuming light travelling from left to right, the image is formed on the right of eyelens at a
distance 2 cm.
Distance of the second principal point P2 from the eyelens L2 is
Negative sign shows that point P2 is on the left of L2 at a distance 4 cm.
Example 4:
The focal length of each lens of a Ramsden’s eyepiece is 4 cm. If light from the sun
is falling upon this eyepiece, locate the positions of the image thus formed and also find the
point from which the distance of the image is to be measured.
Solution:
Equivalent focal length,
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Eye – pieces (Oculars) and their Cardinal Points
Sun is at infinity. Rays coming from sun are parallel. Image will be formed by the eyepiece
at second focal point F2. Hence image is formed at a distance F2 = 3 cm from the second
principal point P2. Distance of second principal point P2 from the second lens (eyelens) L2 is
Second principal point P2 lies at a distance 2 cm to the left of eye – piece.
Distance of the image from eyelens which is the distance of second focal point F 2 from the
eyelens is
Image is formed at a distance of 1 cm to the right of eyelens.
Example 5:
Find lengths of the lenses of Huygens’ eyepiece are 4 cm and 12 cm. Find the
position of cardinal points and plot them in a diagram. If an object is situated 6 cm infront
of the field lens, find the position of the image formed by the eyepiece.
Solution :
In a Huygens’ eyepiece if focal length of field lens (L 1) is f1, then the focal length of
eyelens (L2) is f2 such that
Distance between them is
Focal length of the Huygens’ eyepiece is
Positions of Principal points (P1 and P2)
Distance
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Eye – pieces (Oculars) and their Cardinal Points
P1 is on the right of P
Distance
P2 is on the left of L2
System being in air, P1 and P2 are also nodal points N1 and N2
Position of Focal Points (F1 and F2)
Distance
Distance
F1 is on the right of L1
F2 is on the right of L2
Cardinal points (P1, P2), (F1, F2), (N1, N2) have been shown in Fig. 10
Given object O lies at 6 cm in front of lens L.
L1
L2
N2
Let U and V be the distances of
object O and image I from the first
and second principal points P1 and P2
respectively.
N1 I
P2
O
6cm
F1
F2
P1
5cm
8cm
6cm
2cm
4cm
12cm
9cm
L1O=6cm
Fig.10
Hence,
Thus, the image I is 5 cm behind the eye – lens
Example 6:
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Eye – pieces (Oculars) and their Cardinal Points
From the condition of no chromatic aberration and minimum spherical aberration of a
combination of two separated thin plano-convex lenses, design a combination of equivalent
focal length 6 cm. State the name of this eyepieces. Deduce the position of cardinal points
of this eyepiece and show the path of a pencil of rays through this eye – piece seen by an
eye in relaxed accommodation.
Solution :
Let f1 and f2 be the focal lengths of field lens and eye lens respectively and x be the
distance of their separation.
For no chromatic aberration
For minimum spherical aberration
Focal length of equivalent lens is
Given F = 6 cm
Since
, the given eyepiece is Huygens’ eyepiece.
Cardinal Points:
Principal Points (P1, P2)
P1 lies at 12 cm to the right of field lens L1
P2 lies at 4 cm to the left of eye-lens.
Focal Points (F1 , F2)
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Eye – pieces (Oculars) and their Cardinal Points
Distance of F1 from L1 is
F1 lies at 6 cm to the right of field lens
Distance of second focal point F2 from eyelens L2 is
L1
Second focal point F2 lies at a distance of 2
cm to the right of eye – lens L2.
L2
N2
N1
F1
P2
F2
6cm
2cm
4cm
12cm
8cm
Fig.11(a)
P1
Nodal Points (N1 , N2)
Medium on either side of the combination
being same, nodal points N1 , N2 coincide
with the principal points P1 , P2
respectively. The six cardinal points (P1 ,
P2), (F1 , F2) and (N1 , N2) are shown in
Figs.
11(a).
L1
Path
L2
rays is
eye
shown
in Fig. 11(b)
Position of
equivalent lens
Fig.11(b)
Example 7:
A Ramsden compound eyepiece consists of two thin convex lenses of the same focal
length  f  spaced apart a distance x, where
. Prove that, for relaxed – eye viewing,
the object viewed (formed by the telescope or microscope objective) should be
from the
first lens.
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Eye – pieces (Oculars) and their Cardinal Points
Solution:
Our eye is most relaxed when it is observing objects that are a long distance from
the eye. Thus, for relaxed – eye viewing, the image in the second lens L2 (or eyelens) must
be formed an infinite distance away from the second lens, L 2.
For L2, let the object be situated at a
distance u2 form L2 . Using thin lens
formula, we get
u1
x
v1
This object lies to the left of the first
lens L1.
In fact this is the image
formed by first lens L1 and this acts an
object for lens L2. Thus, the image
formed by the first lens will then be
virtual and at a distance v1 given by
eye
A
B
L1
L2
Fig.12 Example 7
This is shown by B which is the image of object A. Distance of object A from L 1 is u1
This distance is from the first lens L1. Using lens equation, we get
Since
Example 8:
Two converging lenses of focal lengths 4 cm and 5 cm respectively are placed 20 cm
apart and used with the 4 cm lens as the objective and the 5 cm lens as the eyepiece of a
microscope. Where must the object be placed, so that the final image may be 25 cm from
the eye and virtual ?
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5cm
4cm
Eye – pieces (Oculars) and their
Cardinal Points
5.35cm
Solution:
We proceed to solve from the position of
the final image which is at 25 cm from the eye.
v = 25 cm, focal length of eyepiece (L 2), f2 =
5 cm.
L1
L2
20cm
Fig.13
Using thin lens formula,
For the objective (L1)
which gives
where
The negative sign shows that the object be placed before lens L, (which is objective).
Summary
An eye-piece is a spherically designed coaxial system of two lenses, separated by a
distance.
The lens facing the objective is called field lens and the other lens placed near the eye is
called eye-lens. The combination of the two lenses is so designed as to reduce the defect of
images such as spherical aberration, astigmatism, distortion and chromatic aberration.
Function of eye-lens is mainly to magnify the image due to the field lens and to form it at
infinity (for normal vision). Function of field lens is to increase the field of view and to
increase the magnification of the final image seen by the eye.
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Eye – pieces (Oculars) and their Cardinal Points
In order to make some measurements on the final image formed by the eyepiece, we
require a cross-wire or a micrometer scale to be placed coincident with the position of final
image inside the eyepiece. Thus, as the final image formed by the eyepiece is at infinity,
therefore it must lie in the first focal plane of the eyepiece.
The eye-piece is called positive eye-piece when the first focal plane of the eyepiece lies
outside the eyepiece in the object space (i.e. in front of the field lens).
This eyepiece is called Ramsden’s eyepiece. If the first focal plane of the eyepiece lies within
the two lenses of the eyepiece, the final image will be formed on the negative side of the
field lens. Such an eyepiece is called negative eyepiece or Huygens’ eyepiece. Such an
eyepiece cannot be used with cross-wire as the positive eyepiece.
Huygens’ eyepiece is used in low power microscope whereas Ram den’s eyepiece is used in
the spectrometer telescopes. Gauss’ eyepiece is just a Ram den’s eyepiece with a side
opening between the two lenses of the eyepiece and it is used in spectrometer telescopes. It
is useful in focusing the collimator and telescope of spectrometer for infinity.
Exercise
1. What do you understand by an eyepiece? What is its function in an optical
instrument like telescope and microscope. Describe its construction.
2. State difference between an eyepiece and a magnifier.
3. Why does an eyepiece consists of two lenses instead of only one ? What type of
eyepiece is used in an optical instrument ? What is meant by an achromatic
combination of lenses?
4. Explain the importance of plano – convex lenses used in eye – piece. Explain the
difference between the construction and working of Ramsden’s eyepiece and
Huygen’s eyepiece ? Why are cross-wires not fitted in Huygens’ eyepiece ?
5. With neat diagrams discuss the theory and construction of Huygens’ eyepiece and
Ramsden’s eyepiece and trace the course of rays through them. Discuss their
relative merits.
6. (a). State the function of an eyelens and field lens in an eyepiece.
(b). State the position of cross-wires in an eyepiece. Explain its function.
7. A Ramsden’s eyepiece is to have an effective focal length of 3 cm.
focal lengths of the lens components and their distance of separation.
[Ans: 4 cm; 2.67 cm]
Calculate the
8. What are advantages of an eyepiece over a single lens of equivalent focal length ?
Describe the construction, theory and working of Huygens’ eyepiece and show that it
satisfies the conditions of minimum spherical and chromatic aberration.
9. Can Huygens’ eyepiece be used to see our palm ? State with reason.
[Ans: No. The first focal plane of the Huygens’ eyepiece falls within the eyepiece,
where no real object can be placed. This eyepiece may be used only for viewing the
images formed by objectives of microscope and telescope. Ramsden’s eyepiece
alone can be used as amagnifier, its first focal plane lies outside the eyepiece.
10. Describe the construction, theory and uses of a Ramsden’s eyepiece.
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Eye – pieces (Oculars) and their Cardinal Points
11. Deduce the positions of the cardinal points of a Ramsden’s eyepiece and indicate
them on a diagram. Find the position of cross-wire also.
12. Obtain and mark the position of cardinal points of a Huygens’ eyepiece.
position of cross-wires also.
Find the
13. Distinguish between positive and negative eyepiece. With the help of a neat
diagram show that Huygens’ eyepiece is free from both spherical as well as
chromatic aberration.
14. A Ramsden’s eyepiece has equivalent focal length of 4.2 cm.
Deduce its
composition.
[Ans: Plano-convex lens each of focal length 6 cm and their distance of separation
3.8 cm apart]
15. The lenses in a Huygens’ eyepiece have focal length of 2 cm and 4 cm . Find the
distance between the lenses and locate the cardinal points.
[Ans: 3 cm; Cardinal Points L1P1 = L1N1 = 2 cm; L2P2 = L2N2 = 4 cm;
L1F1 = 0.67 cm; L2F2 = 1.33 cm]
16. What
(i)
(ii)
(iii)
type of eyepiece would you recommend for
Spectrometer telescope
Low power microscope
Fresnel’s biprism experiment
17. Discuss optical properties of Ramsden’s eyepiece and Huygens’ eyepiece.
18. Distingush between a magnifier and an eyepiece.
[Ans: If a converging lens, or a lens system, be used to magnify a real object, it is
called a magnifier. If a converging lens, or a lens sysem is used to magnify a real
image (produced by some other optical system before it) it is referred to as an
eyepiece or an ocular.]
19. Why is Huygens’ eyepiece spoken of as a theoretically perfect eyepiece?
[Ans: It is called so because it satisfies the condition for both minimum spherical
aberration and (transverse) achromatism]
20. What is a compound eyepiece? What is its advantage over a single lens? Derive an
expression for the equivalent focal length of a compound eyepiece. In relation to
this expression discuss the eyepieces of Huygens’ and Ramsden explaining clearly
the distinction between them. Also explain why Huygens’ eyepiece cannot be used
for focusing on cross-wires and micrometer scales.
21. The equivalent focal length of an Huygens’ eyepiece is 0 cm. Calculate the focal
length of the field lens. Locate on a diagram the position of focal points of eyepiece.
[Ans: 10 cm]
Objective Questions:
1. Newton’s corpuscular theory of light could not explain the phenomenon:
(a) reflection
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Eye – pieces (Oculars) and their Cardinal Points
(b) refraction
(c) diffraction
(d) rectilinear propagation
Ans.: 1 (c).
The theory could not explain the phenomenon of diffraction which proves that light is a wave
motion.
2. Rectilinear propagation of light is only approximately true because
(a) light consists of quanta (or photons)
(b) light has very large velocity
(c) light has very small wavelength
(d) light exhibits the phenomenon of diffraction
Ans.: 2 (c).
Diffraction of light is bending of light round the corners of the edges of objects. Diffraction is
possible when size of edge is comparable to the wavelength of light used. Due to diffraction
sharp image of any object is not observed because some light inters the geometrical image
(shadow) of the object. Diffraction is almost negligible if the size of the edges is very large as
compared its wavelength of light. Hence light because of its very small wavelength travels
approximately in straight lines.
Fill in the blanks:
3. Monochromatic light of wavelength 6000 Å
in vacuum enters a medium of regractive index 1.5.
_______________ and frequency is ________________.
Its wavelength in the medium is
Ans.: 4000 Å; 51014 Hz
Solution:
4. The speed of light in water is _____________ than in glass, but ______________ than that in
vacuum.
Ans.: more; less.
Speed of light is maximum in vacuum but lies in denser medium and greater in rarer medium.
True/False
5. The speed of light in vacuum depends upon wavelength, frequency and speed of source.
Ans.: False.
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Eye – pieces (Oculars) and their Cardinal Points
The speed of light in vacuum is independent of wavelength, frequency and speed of source; it
remains constant in cacuum.
6. Match the theory of light mentioned in column I with the phenomenon mentioned in column II
Column I
Column II
A
Newton’s corpuscular theory of light
p
Origin of spectra
B
Huygens’s wave theory of light
q
Photoelectric effect
C
Maxwell’s
electromagnetic
wave r
Rectilinear propagation of light
theory of light
D
Quantum of radiation
s
Laws of reflection and refraction
t
Polarisat
Ans.: A  s
Br, s, t
Cr, s, t
Dp, q
b
d
d
Question 7, 8, 9 are based on the paragraph given below. The
figure shows a surface XY separating two transparent media,
Medium
a
c
medium – I and medium II. The lines ab and cd represent
wavefronts of a light wave travelling of a light wave travelling X
I
fh Y
Medium
in medium – I and incident on XY. The lines ef and gh
represent wavefronts of the light wave in medium – II after
e
II
refraction 7. Light travels as a
g
(a) parallel beam in each medium
(b) convergent beam in each medium
(c) divergent beam in each medium
(d) divergent beam in one medium and converging beam in the other medium
Ans (a)
Light ray is defined as a line perpendicular to wavefront. Since wavefronts in each
medium are parallel to each other, light ray travel as a parallel beam in medium.
8. The phases of the light wave at c, d, e and f and
that
,
and
respectively. It is given
(a)
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Eye – pieces (Oculars) and their Cardinal Points
(b)
(c)
(d)
Ans (c)
Points lying on the same wavefront are always the same phase. Hence, phases
and
9. Speed of light is
(a) the same in medium – I and medium – II
(b) larger in medium – I than in medium – II
(c) larger in medium – II than in medium – I
(d) different at b and d
Ans (b)
In terms ray diagram, Angle of incidence 2 is greater than
angle of refraction r . So, medium – II in denser or speed of
light decreases in medium – II.
Alternative Explanation
The perpendicular distance between the wavefronts decreases
as it crosses XY from medium – I to medium – II. Hence,
speed of light decreases from medium – I to medium – II.
Medium I
X
Y
Medium II
Assertion/Reason
The question consists of two statements. Answer the question
on the basis of codes given below:
(a) If statement I is true, statement II is true;
Statement II is the correct explanation for statement I.
(b) Statement is true, statement II is true;
Statement II is NOT a correct explanation for statement I.
(c) If statement I is truel statement II is false
(d) If statement I is false, statement II is true
10. Statement I: The perpendicular distance between the plane wavefronts decreases as it denser
medium.
Statement II: Speed of light is constant and maximum in vacuum.
Ans(a): In figure, AB is incident wavefront and CD is refracted wavefront. By the time the
wavelet from B reach the interface XY at C the wavelet from A will acquire a radius AD of the
wavelet into, medium II, therefore, as time taken by wavelets in same,
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Eye – pieces (Oculars) and their Cardinal Points
A Medium II
Whose c is speed of light in rarer medium I and v is sped of
light in denser medium II. As AD is less than BC so v c
X
speed of light c is maximum in vacuum.
C
A
Medium II
Y
O
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