Ray Optics (Part-2) JEE Notes | EduRev

JEE : Ray Optics (Part-2) JEE Notes | EduRev

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Solution Ans. 1.5
For case (a)  
1
1
v
?
?
?
 = 
1
R
? ?
, 
2 1
1
v v R
?
?
?
=
1 ? ?
?
 and v
2
 = 
2R
m
 ? m=2 ? ? 1 ? ? µ
For case (b) 
2
1
v
?
?
?
= 
1
R
? ?
?
 and v
2
 = 
R
m 1 ?
 ? µ = m
Therefore ? ? 2 1 ? ? ? ? µ ? 
1
1
2
? ? ? ? µ = 1.5
Example#27
In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R
= 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right
of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such
that there is no parallax between final images formed by the lens and mirror. If transverse length of final
image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens
and object.
O
a b
M
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Solution Ans. 5
Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images
formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.
Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image
formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object,
therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a,
therefore, distance of image from L must be equal to 2a.
? (a + 2b) = 2a ? b = 
a
2
The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in
contact with each other co–axially as shown in figure.
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Focal length of convex lens f
1
 is given by  
1
1
f
= ( ?–1) 
1 2
1 1
R R
? ?
?
? ?
? ?
? f
1
 = 40 cm
For concave mirror focal length, f
m
 = 
R
2
= – 20 cm
The combination L behaves like a mirror whose equivalent focal length F is given by
m 1
1 1 2
F f f
? ?
 ? F = –10 cm,
Hence, for the combination u = –a, v = +2a, F = –10 cm
Using mirror formula, 
1 1 1
v u F
? ?
? a = 5 cm
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Solution Ans. 1.5
For case (a)  
1
1
v
?
?
?
 = 
1
R
? ?
, 
2 1
1
v v R
?
?
?
=
1 ? ?
?
 and v
2
 = 
2R
m
 ? m=2 ? ? 1 ? ? µ
For case (b) 
2
1
v
?
?
?
= 
1
R
? ?
?
 and v
2
 = 
R
m 1 ?
 ? µ = m
Therefore ? ? 2 1 ? ? ? ? µ ? 
1
1
2
? ? ? ? µ = 1.5
Example#27
In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R
= 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right
of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such
that there is no parallax between final images formed by the lens and mirror. If transverse length of final
image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens
and object.
O
a b
M
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Solution Ans. 5
Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images
formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.
Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image
formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object,
therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a,
therefore, distance of image from L must be equal to 2a.
? (a + 2b) = 2a ? b = 
a
2
The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in
contact with each other co–axially as shown in figure.
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Focal length of convex lens f
1
 is given by  
1
1
f
= ( ?–1) 
1 2
1 1
R R
? ?
?
? ?
? ?
? f
1
 = 40 cm
For concave mirror focal length, f
m
 = 
R
2
= – 20 cm
The combination L behaves like a mirror whose equivalent focal length F is given by
m 1
1 1 2
F f f
? ?
 ? F = –10 cm,
Hence, for the combination u = –a, v = +2a, F = –10 cm
Using mirror formula, 
1 1 1
v u F
? ?
? a = 5 cm
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Distance of Bird as seen by fish  x
Bf 
= d + ?h
h
d
By differentiating 
fB
d(x ) dh 1 d(d)
dt dt dt
? ?
?
v
FB
 ? 3 + 
4
?
= 6 cm/s
dh d(d) 4
5 2 3 cm / s, 2 ( 2) 4 cm / s,
dt dt 3
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
v
BF
 = 
d(d) dh
dt dt
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 4 + 
4
3
? ?
? ?
? ?
(3) ? 8 cm/s
d(d)
dt
(for fish image after reflection = 0 ) ? 3 + 
1
?
(0)  = 3 cm/s
Similarly speed of image of bird ? 4 cm/s
Example#25
In the shown figure the focal length of equivalent system in the form of 
50x
13
? ?
? ?
? ?
. Find the value of x.
Ans. 2
1
1 3 1 1 1
1
f 2 10 10 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
; 
2
1 6 1 1 3
1
f 5 10 20 100
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 and 
3
1 8 1 1 3
1
f 5 20 20 50
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
1 2 3
1 1 1 1 1 3 3
f f f f 10 100 50
?
? ? ? ? ? ?
=
100
13
=
50x
13
? x=2
Example#26
Quarter part of a transparent cylinder ABC of radius R  is kept on a horizontal floor and a horizontal beam of light
falls on the cylinder in the two different arrangement of cylinder as   shown in the figure (a) & (b). In arrangement
(a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point
E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material.
C B
A
D
2R/m
B C E
R
m -1
(a) (b)
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Solution Ans. 1.5
For case (a)  
1
1
v
?
?
?
 = 
1
R
? ?
, 
2 1
1
v v R
?
?
?
=
1 ? ?
?
 and v
2
 = 
2R
m
 ? m=2 ? ? 1 ? ? µ
For case (b) 
2
1
v
?
?
?
= 
1
R
? ?
?
 and v
2
 = 
R
m 1 ?
 ? µ = m
Therefore ? ? 2 1 ? ? ? ? µ ? 
1
1
2
? ? ? ? µ = 1.5
Example#27
In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R
= 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right
of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such
that there is no parallax between final images formed by the lens and mirror. If transverse length of final
image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens
and object.
O
a b
M
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Solution Ans. 5
Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images
formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.
Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image
formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object,
therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a,
therefore, distance of image from L must be equal to 2a.
? (a + 2b) = 2a ? b = 
a
2
The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in
contact with each other co–axially as shown in figure.
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Focal length of convex lens f
1
 is given by  
1
1
f
= ( ?–1) 
1 2
1 1
R R
? ?
?
? ?
? ?
? f
1
 = 40 cm
For concave mirror focal length, f
m
 = 
R
2
= – 20 cm
The combination L behaves like a mirror whose equivalent focal length F is given by
m 1
1 1 2
F f f
? ?
 ? F = –10 cm,
Hence, for the combination u = –a, v = +2a, F = –10 cm
Using mirror formula, 
1 1 1
v u F
? ?
? a = 5 cm
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Distance of Bird as seen by fish  x
Bf 
= d + ?h
h
d
By differentiating 
fB
d(x ) dh 1 d(d)
dt dt dt
? ?
?
v
FB
 ? 3 + 
4
?
= 6 cm/s
dh d(d) 4
5 2 3 cm / s, 2 ( 2) 4 cm / s,
dt dt 3
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
v
BF
 = 
d(d) dh
dt dt
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 4 + 
4
3
? ?
? ?
? ?
(3) ? 8 cm/s
d(d)
dt
(for fish image after reflection = 0 ) ? 3 + 
1
?
(0)  = 3 cm/s
Similarly speed of image of bird ? 4 cm/s
Example#25
In the shown figure the focal length of equivalent system in the form of 
50x
13
? ?
? ?
? ?
. Find the value of x.
Ans. 2
1
1 3 1 1 1
1
f 2 10 10 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
; 
2
1 6 1 1 3
1
f 5 10 20 100
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 and 
3
1 8 1 1 3
1
f 5 20 20 50
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
1 2 3
1 1 1 1 1 3 3
f f f f 10 100 50
?
? ? ? ? ? ?
=
100
13
=
50x
13
? x=2
Example#26
Quarter part of a transparent cylinder ABC of radius R  is kept on a horizontal floor and a horizontal beam of light
falls on the cylinder in the two different arrangement of cylinder as   shown in the figure (a) & (b). In arrangement
(a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point
E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material.
C B
A
D
2R/m
B C E
R
m -1
(a) (b)
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Example#23
Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image.
Column I Column II
(A)
F C
(P) real, inverted, enlarged
(B) (Q) virtual, erect, enlarged
(C) (R) virtual, erect, diminished
(D)
F
(S) virtual, erect
Solution Ans. (A) P (b)RS (C) S (D) QS
Example#24
A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is   silvered. A  fish in the tank is rising
upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ?
water
 = 4/3]
Column I (cm/s) Column II
(A) Speed of the image of fish as seen by the bird directly (P) 8
(B) Speed of the image of fish formed after reflection in (Q) 6
the mirror as seen by the bird
(C) Speed of image of bird relative to the fish looking upwards (R) 3
(D) Speed of image of bird relative to the fish looking (S) 4
downwards in the mirror
Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S)
Distance of fish asseen by bird  x
fB
 = h + 
d
?
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Solution Ans. 1.5
For case (a)  
1
1
v
?
?
?
 = 
1
R
? ?
, 
2 1
1
v v R
?
?
?
=
1 ? ?
?
 and v
2
 = 
2R
m
 ? m=2 ? ? 1 ? ? µ
For case (b) 
2
1
v
?
?
?
= 
1
R
? ?
?
 and v
2
 = 
R
m 1 ?
 ? µ = m
Therefore ? ? 2 1 ? ? ? ? µ ? 
1
1
2
? ? ? ? µ = 1.5
Example#27
In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R
= 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right
of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such
that there is no parallax between final images formed by the lens and mirror. If transverse length of final
image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens
and object.
O
a b
M
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Solution Ans. 5
Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images
formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.
Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image
formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object,
therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a,
therefore, distance of image from L must be equal to 2a.
? (a + 2b) = 2a ? b = 
a
2
The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in
contact with each other co–axially as shown in figure.
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Focal length of convex lens f
1
 is given by  
1
1
f
= ( ?–1) 
1 2
1 1
R R
? ?
?
? ?
? ?
? f
1
 = 40 cm
For concave mirror focal length, f
m
 = 
R
2
= – 20 cm
The combination L behaves like a mirror whose equivalent focal length F is given by
m 1
1 1 2
F f f
? ?
 ? F = –10 cm,
Hence, for the combination u = –a, v = +2a, F = –10 cm
Using mirror formula, 
1 1 1
v u F
? ?
? a = 5 cm
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Distance of Bird as seen by fish  x
Bf 
= d + ?h
h
d
By differentiating 
fB
d(x ) dh 1 d(d)
dt dt dt
? ?
?
v
FB
 ? 3 + 
4
?
= 6 cm/s
dh d(d) 4
5 2 3 cm / s, 2 ( 2) 4 cm / s,
dt dt 3
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
v
BF
 = 
d(d) dh
dt dt
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 4 + 
4
3
? ?
? ?
? ?
(3) ? 8 cm/s
d(d)
dt
(for fish image after reflection = 0 ) ? 3 + 
1
?
(0)  = 3 cm/s
Similarly speed of image of bird ? 4 cm/s
Example#25
In the shown figure the focal length of equivalent system in the form of 
50x
13
? ?
? ?
? ?
. Find the value of x.
Ans. 2
1
1 3 1 1 1
1
f 2 10 10 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
; 
2
1 6 1 1 3
1
f 5 10 20 100
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 and 
3
1 8 1 1 3
1
f 5 20 20 50
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
1 2 3
1 1 1 1 1 3 3
f f f f 10 100 50
?
? ? ? ? ? ?
=
100
13
=
50x
13
? x=2
Example#26
Quarter part of a transparent cylinder ABC of radius R  is kept on a horizontal floor and a horizontal beam of light
falls on the cylinder in the two different arrangement of cylinder as   shown in the figure (a) & (b). In arrangement
(a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point
E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material.
C B
A
D
2R/m
B C E
R
m -1
(a) (b)
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Example#23
Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image.
Column I Column II
(A)
F C
(P) real, inverted, enlarged
(B) (Q) virtual, erect, enlarged
(C) (R) virtual, erect, diminished
(D)
F
(S) virtual, erect
Solution Ans. (A) P (b)RS (C) S (D) QS
Example#24
A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is   silvered. A  fish in the tank is rising
upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ?
water
 = 4/3]
Column I (cm/s) Column II
(A) Speed of the image of fish as seen by the bird directly (P) 8
(B) Speed of the image of fish formed after reflection in (Q) 6
the mirror as seen by the bird
(C) Speed of image of bird relative to the fish looking upwards (R) 3
(D) Speed of image of bird relative to the fish looking (S) 4
downwards in the mirror
Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S)
Distance of fish asseen by bird  x
fB
 = h + 
d
?
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Example#19 to 21
There is a spherical glass ball of refractive index ?
1
 and another glass ball of refractive index ?
2
 inside it
as shown in figure. The radius of the outer ball is R
1
 and that of inner ball is R
2
. A ray is incident on the
outer surface of the ball at an angle i
1
.
? 1
? 2
R
1
r 1
i 2
B
C
O
R 2
A
i 1
D
r 2
1 9 . Find the value of r
1
(A) 
1 1
1
sin i
sin
?
? ?
? ?
? ? ?
(B) ? ?
1
1 1
sin sin i
?
? (C) 
1 1
1
sin 
sin i
?
? ? ?
? ?
? ?
(D) 
1
1 1
1
sin 
sin i
?
? ?
? ?
? ? ?
2 0 . Find the value of i
2
(A) 
1 2 1
1 1
R sin i
sin 
R
?
? ?
? ?
? ? ?
(B) ]
1 1 1
2 2
R sin i
sin 
R
?
? ?
? ?
? ? ?
(C) 
1 1 1
2 1
R sin i
sin 
R
?
? ?
? ?
? ? ?
(D) 
1 2 1
1 2
R sin i
sin 
R
?
? ?
? ?
? ? ?
2 1 . Find the value of r
2
(A) 
1 1
1
2 2
R
sin sin i
R
?
? ?
? ?
? ? ?
(B) 
1 2
1
2 1
R
sin sin i
R
?
? ?
? ?
? ? ?
(C) 
1 1
1 2 1
R 1
sin 
R sin i
?
? ?
? ?
? ? ?
(D) 
1 2
1
1 1
R
sin sin i
R
?
? ?
? ?
? ? ?
Solution
1 9 . Ans. (A)
?
1
 sin r
1
 = sin i
1 
? r
1
 = 
1 1
1
sin i
sin
?
? ?
? ?
? ? ?
2 0 . Ans. (C)
Using sine rule 
?
?
1 2
2 1
sin r sin(180 i )
R R
? ??sin i
2
 = 
1
2
R
R
 sin r
1
 = 
1 1
2 1
R sin i
R
? ?
? ?
? ? ?
1 1 1
2
2 1
R sin i
i sin
R
?
? ?
? ?
? ?
? ? ?
2 1 . Ans. (A)
1
2
?
?
 sin i
2
 = sin r
2
; 
1 1 1
2 2 1
R sin i
R
?
? ?
= sin r
2
; r
2
 = 
1 1
1
2 2
R
sin sin i
R
?
? ?
? ?
? ? ?
Example#22
Consider a an equilateral prism ABC of glass 
3
2
? ?
? ?
? ?
? ?
 placed in water 
4
3
? ?
? ?
? ?
? ?
A
water
=4/3 ?
glass =3/2 ?
Column-I Column-II
(A) FG is parallel to BC (P) Maximum deviation
(B) i
1
 = 90
0
(Q) Minimum deviation
(C) i
1
 = i
2
 = sin
-1
 
9
16
? ?
? ?
? ?
(R) TIR will take place at surface AC
(D) EF is perpendicular to AB (S) No TIR will take place at surface BC
Solution Ans. (A) QS, (B) PS, (C) QS, (D) S
At minimum deviation i
1
 = i
2
, EF   BC;          At maximum deviation i
1
 = 90
0
 or i
2
=90
0
For i
1
 =0, TIR will not take place at AC
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Solution Ans. 1.5
For case (a)  
1
1
v
?
?
?
 = 
1
R
? ?
, 
2 1
1
v v R
?
?
?
=
1 ? ?
?
 and v
2
 = 
2R
m
 ? m=2 ? ? 1 ? ? µ
For case (b) 
2
1
v
?
?
?
= 
1
R
? ?
?
 and v
2
 = 
R
m 1 ?
 ? µ = m
Therefore ? ? 2 1 ? ? ? ? µ ? 
1
1
2
? ? ? ? µ = 1.5
Example#27
In figure, L is half part of an equiconvex glass lens ( ? = 1.5) whose surfaces have radius of curvature R
= 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right
of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such
that there is no parallax between final images formed by the lens and mirror. If transverse length of final
image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens
and object.
O
a b
M
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Solution Ans. 5
Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images
formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.
Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image
formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object,
therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a,
therefore, distance of image from L must be equal to 2a.
? (a + 2b) = 2a ? b = 
a
2
The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in
contact with each other co–axially as shown in figure.
\ \ \ \ \ \ \ \ \ \ \ \ \ \
Focal length of convex lens f
1
 is given by  
1
1
f
= ( ?–1) 
1 2
1 1
R R
? ?
?
? ?
? ?
? f
1
 = 40 cm
For concave mirror focal length, f
m
 = 
R
2
= – 20 cm
The combination L behaves like a mirror whose equivalent focal length F is given by
m 1
1 1 2
F f f
? ?
 ? F = –10 cm,
Hence, for the combination u = –a, v = +2a, F = –10 cm
Using mirror formula, 
1 1 1
v u F
? ?
? a = 5 cm
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Distance of Bird as seen by fish  x
Bf 
= d + ?h
h
d
By differentiating 
fB
d(x ) dh 1 d(d)
dt dt dt
? ?
?
v
FB
 ? 3 + 
4
?
= 6 cm/s
dh d(d) 4
5 2 3 cm / s, 2 ( 2) 4 cm / s,
dt dt 3
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
v
BF
 = 
d(d) dh
dt dt
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 4 + 
4
3
? ?
? ?
? ?
(3) ? 8 cm/s
d(d)
dt
(for fish image after reflection = 0 ) ? 3 + 
1
?
(0)  = 3 cm/s
Similarly speed of image of bird ? 4 cm/s
Example#25
In the shown figure the focal length of equivalent system in the form of 
50x
13
? ?
? ?
? ?
. Find the value of x.
Ans. 2
1
1 3 1 1 1
1
f 2 10 10 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
; 
2
1 6 1 1 3
1
f 5 10 20 100
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 and 
3
1 8 1 1 3
1
f 5 20 20 50
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
1 2 3
1 1 1 1 1 3 3
f f f f 10 100 50
?
? ? ? ? ? ?
=
100
13
=
50x
13
? x=2
Example#26
Quarter part of a transparent cylinder ABC of radius R  is kept on a horizontal floor and a horizontal beam of light
falls on the cylinder in the two different arrangement of cylinder as   shown in the figure (a) & (b). In arrangement
(a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point
E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material.
C B
A
D
2R/m
B C E
R
m -1
(a) (b)
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Example#23
Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image.
Column I Column II
(A)
F C
(P) real, inverted, enlarged
(B) (Q) virtual, erect, enlarged
(C) (R) virtual, erect, diminished
(D)
F
(S) virtual, erect
Solution Ans. (A) P (b)RS (C) S (D) QS
Example#24
A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is   silvered. A  fish in the tank is rising
upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : ?
water
 = 4/3]
Column I (cm/s) Column II
(A) Speed of the image of fish as seen by the bird directly (P) 8
(B) Speed of the image of fish formed after reflection in (Q) 6
the mirror as seen by the bird
(C) Speed of image of bird relative to the fish looking upwards (R) 3
(D) Speed of image of bird relative to the fish looking (S) 4
downwards in the mirror
Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S)
Distance of fish asseen by bird  x
fB
 = h + 
d
?
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Example#19 to 21
There is a spherical glass ball of refractive index ?
1
 and another glass ball of refractive index ?
2
 inside it
as shown in figure. The radius of the outer ball is R
1
 and that of inner ball is R
2
. A ray is incident on the
outer surface of the ball at an angle i
1
.
? 1
? 2
R
1
r 1
i 2
B
C
O
R 2
A
i 1
D
r 2
1 9 . Find the value of r
1
(A) 
1 1
1
sin i
sin
?
? ?
? ?
? ? ?
(B) ? ?
1
1 1
sin sin i
?
? (C) 
1 1
1
sin 
sin i
?
? ? ?
? ?
? ?
(D) 
1
1 1
1
sin 
sin i
?
? ?
? ?
? ? ?
2 0 . Find the value of i
2
(A) 
1 2 1
1 1
R sin i
sin 
R
?
? ?
? ?
? ? ?
(B) ]
1 1 1
2 2
R sin i
sin 
R
?
? ?
? ?
? ? ?
(C) 
1 1 1
2 1
R sin i
sin 
R
?
? ?
? ?
? ? ?
(D) 
1 2 1
1 2
R sin i
sin 
R
?
? ?
? ?
? ? ?
2 1 . Find the value of r
2
(A) 
1 1
1
2 2
R
sin sin i
R
?
? ?
? ?
? ? ?
(B) 
1 2
1
2 1
R
sin sin i
R
?
? ?
? ?
? ? ?
(C) 
1 1
1 2 1
R 1
sin 
R sin i
?
? ?
? ?
? ? ?
(D) 
1 2
1
1 1
R
sin sin i
R
?
? ?
? ?
? ? ?
Solution
1 9 . Ans. (A)
?
1
 sin r
1
 = sin i
1 
? r
1
 = 
1 1
1
sin i
sin
?
? ?
? ?
? ? ?
2 0 . Ans. (C)
Using sine rule 
?
?
1 2
2 1
sin r sin(180 i )
R R
? ??sin i
2
 = 
1
2
R
R
 sin r
1
 = 
1 1
2 1
R sin i
R
? ?
? ?
? ? ?
1 1 1
2
2 1
R sin i
i sin
R
?
? ?
? ?
? ?
? ? ?
2 1 . Ans. (A)
1
2
?
?
 sin i
2
 = sin r
2
; 
1 1 1
2 2 1
R sin i
R
?
? ?
= sin r
2
; r
2
 = 
1 1
1
2 2
R
sin sin i
R
?
? ?
? ?
? ? ?
Example#22
Consider a an equilateral prism ABC of glass 
3
2
? ?
? ?
? ?
? ?
 placed in water 
4
3
? ?
? ?
? ?
? ?
A
water
=4/3 ?
glass =3/2 ?
Column-I Column-II
(A) FG is parallel to BC (P) Maximum deviation
(B) i
1
 = 90
0
(Q) Minimum deviation
(C) i
1
 = i
2
 = sin
-1
 
9
16
? ?
? ?
? ?
(R) TIR will take place at surface AC
(D) EF is perpendicular to AB (S) No TIR will take place at surface BC
Solution Ans. (A) QS, (B) PS, (C) QS, (D) S
At minimum deviation i
1
 = i
2
, EF   BC;          At maximum deviation i
1
 = 90
0
 or i
2
=90
0
For i
1
 =0, TIR will not take place at AC
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Solution
13. Ans. (A)
Time t =
distance
speed
=
? ?
2
2 2 2
1 2
Y x x Y
c
? ? ? ? ?
1 4 . Ans. (A)
For least time 
dt
dx
=0 ?
2 2
1
2x
Y x ?
–
? ?
? ?
2
2
2
2 x
x Y
?
? ?
?
?
 = 0 ?
1 2
sin sin ? ? ? ?
1 2
? ? ?
1 5 . Ans. (A)
Example#16 to 18
One hard and stormy night you find yourself lost in the forest when you come upon a small hut. Entering it you
see a crooked old woman in the corner hunched over a crystal ball. You are about to make a hasty exit when
you hear the howl of wolves outside. Taking another look at the gypsy you decide to take your chances with the
wolves, but the door is jammed shut. Resigned to a bad situation you approach her slowly, wondering just what
is the focal length of that nifty crystal ball.
1 6 . If the crystal ball is 20 cm in diameter with R.I. = 1.5, the gypsy lady is 1.2 m from the ball, where is
the image of the gypsy in focus as you walk towards her?
(A) 6.9 cm from the crystal ball (B) 7.9 cm from the crystal ball
(C) 8.9 cm from the crystal ball (D) None of these
1 7 . The image of old lady is
(A) real, inverted and enlarged (B) erect, virtual and small
(C) erect, virtual and magnified (D) real, inverted and small
1 8 . The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no longer get
an image of her. At what object distance will there be no change of the gypsy formed?
(A) 10cm (B) 5 cm (C) 15 cm (D) None of these
Solution
1 6 . Ans. (A)
For refraction at 1
st
 surface 1
1
1.5 1 1.5 1
v 36cm
v 120 10
?
? ? ? ?
? ?
              
Lady
120cm R=10cm
(1)
(2)
for refraction at 2
nd
 surface 
1 1.5 1 1.5 80
v 6.9cm
v (36 20) 10 11.5
?
? ? ? ? ?
? ?
1 7 . Ans. (D)
Total magnification = m
1
m
2
 = 
1 1 2
2 1 1
v v v
u v u
? ? ? ? ? ?
?
? ? ? ?
? ? ? ? ? ?
 = negative e
1 8 . Ans. (B)
At this point image will formed at infinity
For refraction at second surface 1
1
1 1.5 1 1.5
v 30cm
v 10
?
? ? ? ? ?
? ?
            
x
20cm
(1)
(2)
For refraction at first surface 
1.5 1 1.5 1
x 5cm
10 x 10
?
? ? ? ?
? ? ?
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