Refraction of Light (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Refraction of Light (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

 Page 1


Introductory Exercise 28.3 
Q 1.  When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from 
the lens. The lens is made of a material of refractive index and its two spherical surfaces 
have the same radius of curvature. What is the value of this radius? 
Q 2.  A converging lens has a focal length of 30 cm, Rays from a 2.0 cm high filament that pass through 
the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? 
What is the height of the image? 
Q 3.  Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and 
the right surfaces are interchanged. 
Q 4.  As an object is moved from the surface of a thin converging lens to a focal point, over what range 
does the image distance vary? 
Q 5.  A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces 
of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. 
  (a) What is now the focal length of the lens?  
(b) What is the minimum distance that an immersed object must be from the lens so that a real 
image is formed? 
Q 6.  An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens 
is placed to the right of the first lens and then moved until the image it produces is identical in size 
and orientation to the object. What is the separation between the lenses. 
Q 7.  Suppose an object has thickness du so that it extends from object distance u to u + du.  
  Prove that the thickness dv of its image is given by 
2
2
v
du
u
??
?
??
??
,
 
so the longitudinal magnification 
dv
du
 = -m
2
, where m is the lateral magnification. 
Q 8.  Two thin similar convex glass pieces are joined together front to front, with its rear portion 
silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between the 
glass pieces is replaced by water 
4
3
??
??
??
??
.
 
find the position of image. 
Q 9.  When a pin is moved along the principal axis of a small concave mirror, the image position 
coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 
m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the 
mirror. Find the refractive index of the liquid. 
 
Page 2


Introductory Exercise 28.3 
Q 1.  When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from 
the lens. The lens is made of a material of refractive index and its two spherical surfaces 
have the same radius of curvature. What is the value of this radius? 
Q 2.  A converging lens has a focal length of 30 cm, Rays from a 2.0 cm high filament that pass through 
the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? 
What is the height of the image? 
Q 3.  Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and 
the right surfaces are interchanged. 
Q 4.  As an object is moved from the surface of a thin converging lens to a focal point, over what range 
does the image distance vary? 
Q 5.  A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces 
of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. 
  (a) What is now the focal length of the lens?  
(b) What is the minimum distance that an immersed object must be from the lens so that a real 
image is formed? 
Q 6.  An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens 
is placed to the right of the first lens and then moved until the image it produces is identical in size 
and orientation to the object. What is the separation between the lenses. 
Q 7.  Suppose an object has thickness du so that it extends from object distance u to u + du.  
  Prove that the thickness dv of its image is given by 
2
2
v
du
u
??
?
??
??
,
 
so the longitudinal magnification 
dv
du
 = -m
2
, where m is the lateral magnification. 
Q 8.  Two thin similar convex glass pieces are joined together front to front, with its rear portion 
silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between the 
glass pieces is replaced by water 
4
3
??
??
??
??
.
 
find the position of image. 
Q 9.  When a pin is moved along the principal axis of a small concave mirror, the image position 
coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 
m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the 
mirror. Find the refractive index of the liquid. 
 
Q 10.  When a lens is inserted between an object and a screen which are a fixed distance apart the size of 
the image is either 6 cm or 
2
3
cm. Find size of the object. 
Q 11.  A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the 
distance in cm between the object and image. 
Q 12.  The distance between an object and its upright image is 20 cm. If the magnification is 0.5, what is 
the focal length of the lens that is being used to form the image? 
Q 13.  A thin lens of focal length + 10.0 cm lies on a horizontal plane mirror. How far above the lens 
should an object be held if its image is to coincide with the object? 
Solutions 
1.    
   
  Solving we get, R = 39 cm  
2.  
  Solving we get u = - 18.75 cm 
    
   I = 
m
(O) = 2.67 × 2 = 5.33cm 
3.  
    
  Solving these two equations we can see that f
1
 = f
2
. 
4.  
When object is moved from O to F
1
 its virtual, erect and magnified image should vary from O to -
?. 
5.  (a) 
   f = +36cm 
  (b) Between O and F
1
 image is virtual. Hence for real image. 
   | ? | < f or 36 cm 
6.  
Page 3


Introductory Exercise 28.3 
Q 1.  When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from 
the lens. The lens is made of a material of refractive index and its two spherical surfaces 
have the same radius of curvature. What is the value of this radius? 
Q 2.  A converging lens has a focal length of 30 cm, Rays from a 2.0 cm high filament that pass through 
the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? 
What is the height of the image? 
Q 3.  Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and 
the right surfaces are interchanged. 
Q 4.  As an object is moved from the surface of a thin converging lens to a focal point, over what range 
does the image distance vary? 
Q 5.  A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces 
of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. 
  (a) What is now the focal length of the lens?  
(b) What is the minimum distance that an immersed object must be from the lens so that a real 
image is formed? 
Q 6.  An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens 
is placed to the right of the first lens and then moved until the image it produces is identical in size 
and orientation to the object. What is the separation between the lenses. 
Q 7.  Suppose an object has thickness du so that it extends from object distance u to u + du.  
  Prove that the thickness dv of its image is given by 
2
2
v
du
u
??
?
??
??
,
 
so the longitudinal magnification 
dv
du
 = -m
2
, where m is the lateral magnification. 
Q 8.  Two thin similar convex glass pieces are joined together front to front, with its rear portion 
silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between the 
glass pieces is replaced by water 
4
3
??
??
??
??
.
 
find the position of image. 
Q 9.  When a pin is moved along the principal axis of a small concave mirror, the image position 
coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 
m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the 
mirror. Find the refractive index of the liquid. 
 
Q 10.  When a lens is inserted between an object and a screen which are a fixed distance apart the size of 
the image is either 6 cm or 
2
3
cm. Find size of the object. 
Q 11.  A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the 
distance in cm between the object and image. 
Q 12.  The distance between an object and its upright image is 20 cm. If the magnification is 0.5, what is 
the focal length of the lens that is being used to form the image? 
Q 13.  A thin lens of focal length + 10.0 cm lies on a horizontal plane mirror. How far above the lens 
should an object be held if its image is to coincide with the object? 
Solutions 
1.    
   
  Solving we get, R = 39 cm  
2.  
  Solving we get u = - 18.75 cm 
    
   I = 
m
(O) = 2.67 × 2 = 5.33cm 
3.  
    
  Solving these two equations we can see that f
1
 = f
2
. 
4.  
When object is moved from O to F
1
 its virtual, erect and magnified image should vary from O to -
?. 
5.  (a) 
   f = +36cm 
  (b) Between O and F
1
 image is virtual. Hence for real image. 
   | ? | < f or 36 cm 
6.  
7.  
  Differentiating this equation, we get, 
   - v
-2
 - dv+ u
-2 
. du = 0  (as f = constant) 
 
8.  It is just like a concave mirror 
  |f| = 0.2m |R|=0.4m Focal length
 
of this equivalent mirror is   
        
   
  or  F= — 0.12 m or -12 cm 
9.  | R | = 0.5 m (from first case) 
  In the shown figure, object appears at distance 
  d= u
e
(0.2) + 0.2  
Now, for image to further coincide with the object, 
   d = |R| Solving we get, ?
e 
= 1.5 
10. 
12
O I I ? (Displacement method) 
    
   = 2 cm 
11.  Virtual, magnified and erect image is formed by convex lens. 
  Let   u = -x 
  Then   v =-3x 
  Now,  
   x = 8 cm 
  Distance between object and image = 3x - x = 2x = 16 cm 
12.  Diminished erect image is formed by concave lens.  
  Let   u = - x then 
x
v
2
??
 
  Now,  |u| - |v| = 20 cm 
x
2
 =20 cm 
  or   x = 40 cm 
  u = -40cm  and   v = -20cm 
    
  or   f = -40cm 
Page 4


Introductory Exercise 28.3 
Q 1.  When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from 
the lens. The lens is made of a material of refractive index and its two spherical surfaces 
have the same radius of curvature. What is the value of this radius? 
Q 2.  A converging lens has a focal length of 30 cm, Rays from a 2.0 cm high filament that pass through 
the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? 
What is the height of the image? 
Q 3.  Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and 
the right surfaces are interchanged. 
Q 4.  As an object is moved from the surface of a thin converging lens to a focal point, over what range 
does the image distance vary? 
Q 5.  A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces 
of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. 
  (a) What is now the focal length of the lens?  
(b) What is the minimum distance that an immersed object must be from the lens so that a real 
image is formed? 
Q 6.  An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens 
is placed to the right of the first lens and then moved until the image it produces is identical in size 
and orientation to the object. What is the separation between the lenses. 
Q 7.  Suppose an object has thickness du so that it extends from object distance u to u + du.  
  Prove that the thickness dv of its image is given by 
2
2
v
du
u
??
?
??
??
,
 
so the longitudinal magnification 
dv
du
 = -m
2
, where m is the lateral magnification. 
Q 8.  Two thin similar convex glass pieces are joined together front to front, with its rear portion 
silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between the 
glass pieces is replaced by water 
4
3
??
??
??
??
.
 
find the position of image. 
Q 9.  When a pin is moved along the principal axis of a small concave mirror, the image position 
coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 
m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the 
mirror. Find the refractive index of the liquid. 
 
Q 10.  When a lens is inserted between an object and a screen which are a fixed distance apart the size of 
the image is either 6 cm or 
2
3
cm. Find size of the object. 
Q 11.  A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the 
distance in cm between the object and image. 
Q 12.  The distance between an object and its upright image is 20 cm. If the magnification is 0.5, what is 
the focal length of the lens that is being used to form the image? 
Q 13.  A thin lens of focal length + 10.0 cm lies on a horizontal plane mirror. How far above the lens 
should an object be held if its image is to coincide with the object? 
Solutions 
1.    
   
  Solving we get, R = 39 cm  
2.  
  Solving we get u = - 18.75 cm 
    
   I = 
m
(O) = 2.67 × 2 = 5.33cm 
3.  
    
  Solving these two equations we can see that f
1
 = f
2
. 
4.  
When object is moved from O to F
1
 its virtual, erect and magnified image should vary from O to -
?. 
5.  (a) 
   f = +36cm 
  (b) Between O and F
1
 image is virtual. Hence for real image. 
   | ? | < f or 36 cm 
6.  
7.  
  Differentiating this equation, we get, 
   - v
-2
 - dv+ u
-2 
. du = 0  (as f = constant) 
 
8.  It is just like a concave mirror 
  |f| = 0.2m |R|=0.4m Focal length
 
of this equivalent mirror is   
        
   
  or  F= — 0.12 m or -12 cm 
9.  | R | = 0.5 m (from first case) 
  In the shown figure, object appears at distance 
  d= u
e
(0.2) + 0.2  
Now, for image to further coincide with the object, 
   d = |R| Solving we get, ?
e 
= 1.5 
10. 
12
O I I ? (Displacement method) 
    
   = 2 cm 
11.  Virtual, magnified and erect image is formed by convex lens. 
  Let   u = -x 
  Then   v =-3x 
  Now,  
   x = 8 cm 
  Distance between object and image = 3x - x = 2x = 16 cm 
12.  Diminished erect image is formed by concave lens.  
  Let   u = - x then 
x
v
2
??
 
  Now,  |u| - |v| = 20 cm 
x
2
 =20 cm 
  or   x = 40 cm 
  u = -40cm  and   v = -20cm 
    
  or   f = -40cm 
13.  If object is placed at focus of lens (= 10 cm), rays become parallel and fall normal on plane mirror. 
So, rays retrace their path. 
Introductory Exercise 28.4 
Q 1.  The prism shown in figure has a refractive index of 1.60 and the angles A are 30°. Two light rays 
P and Q are parallel as they enter the prism. What is the angle between them after they emerge? 
  [sin
-1
 (0.8) = 53°] 
? 
Q 2.  Light is incident normally on the short face of a 30° - 60° - 90° prism. A liquid is poured on the 
hypotenuse of the prism. If the refractive index of the prism is V3, find the maximum refractive 
index of the liquid so that light is totally reflected. 
 
Q 3.  A glass vessel in the shape of a triangular prism is filled with water, and light is incident normally 
on the face XY. If the refractive indices for water and glass are 4/3 and 3/2 respectively, total 
internal reflection will occur at the glass-air surface XZ only for sin 6 greater than 
 
  A  1/2  B  2/3  C  3/4 D  8/9  E  16/27. 
Q 4.  A parallel beam of light is incident on a prism shown in figure. Through what angle should the 
mirror be rotated so that light returns back to its original path? Refractive index of prism is 1.5. 
 
Q 5.  A light ray going through a prism with the angle of prism 60°, is found to deviate at least by 30°. 
What is the range of the refractive index of the prism? 
Q 6.  A ray of light falls normally on a refracting face of a prism. Find the angle of prism if the ray just 
fails to emerge from the prism ( ? = 3/2).  
Page 5


Introductory Exercise 28.3 
Q 1.  When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from 
the lens. The lens is made of a material of refractive index and its two spherical surfaces 
have the same radius of curvature. What is the value of this radius? 
Q 2.  A converging lens has a focal length of 30 cm, Rays from a 2.0 cm high filament that pass through 
the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? 
What is the height of the image? 
Q 3.  Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and 
the right surfaces are interchanged. 
Q 4.  As an object is moved from the surface of a thin converging lens to a focal point, over what range 
does the image distance vary? 
Q 5.  A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces 
of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. 
  (a) What is now the focal length of the lens?  
(b) What is the minimum distance that an immersed object must be from the lens so that a real 
image is formed? 
Q 6.  An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens 
is placed to the right of the first lens and then moved until the image it produces is identical in size 
and orientation to the object. What is the separation between the lenses. 
Q 7.  Suppose an object has thickness du so that it extends from object distance u to u + du.  
  Prove that the thickness dv of its image is given by 
2
2
v
du
u
??
?
??
??
,
 
so the longitudinal magnification 
dv
du
 = -m
2
, where m is the lateral magnification. 
Q 8.  Two thin similar convex glass pieces are joined together front to front, with its rear portion 
silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between the 
glass pieces is replaced by water 
4
3
??
??
??
??
.
 
find the position of image. 
Q 9.  When a pin is moved along the principal axis of a small concave mirror, the image position 
coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 
m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the 
mirror. Find the refractive index of the liquid. 
 
Q 10.  When a lens is inserted between an object and a screen which are a fixed distance apart the size of 
the image is either 6 cm or 
2
3
cm. Find size of the object. 
Q 11.  A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the 
distance in cm between the object and image. 
Q 12.  The distance between an object and its upright image is 20 cm. If the magnification is 0.5, what is 
the focal length of the lens that is being used to form the image? 
Q 13.  A thin lens of focal length + 10.0 cm lies on a horizontal plane mirror. How far above the lens 
should an object be held if its image is to coincide with the object? 
Solutions 
1.    
   
  Solving we get, R = 39 cm  
2.  
  Solving we get u = - 18.75 cm 
    
   I = 
m
(O) = 2.67 × 2 = 5.33cm 
3.  
    
  Solving these two equations we can see that f
1
 = f
2
. 
4.  
When object is moved from O to F
1
 its virtual, erect and magnified image should vary from O to -
?. 
5.  (a) 
   f = +36cm 
  (b) Between O and F
1
 image is virtual. Hence for real image. 
   | ? | < f or 36 cm 
6.  
7.  
  Differentiating this equation, we get, 
   - v
-2
 - dv+ u
-2 
. du = 0  (as f = constant) 
 
8.  It is just like a concave mirror 
  |f| = 0.2m |R|=0.4m Focal length
 
of this equivalent mirror is   
        
   
  or  F= — 0.12 m or -12 cm 
9.  | R | = 0.5 m (from first case) 
  In the shown figure, object appears at distance 
  d= u
e
(0.2) + 0.2  
Now, for image to further coincide with the object, 
   d = |R| Solving we get, ?
e 
= 1.5 
10. 
12
O I I ? (Displacement method) 
    
   = 2 cm 
11.  Virtual, magnified and erect image is formed by convex lens. 
  Let   u = -x 
  Then   v =-3x 
  Now,  
   x = 8 cm 
  Distance between object and image = 3x - x = 2x = 16 cm 
12.  Diminished erect image is formed by concave lens.  
  Let   u = - x then 
x
v
2
??
 
  Now,  |u| - |v| = 20 cm 
x
2
 =20 cm 
  or   x = 40 cm 
  u = -40cm  and   v = -20cm 
    
  or   f = -40cm 
13.  If object is placed at focus of lens (= 10 cm), rays become parallel and fall normal on plane mirror. 
So, rays retrace their path. 
Introductory Exercise 28.4 
Q 1.  The prism shown in figure has a refractive index of 1.60 and the angles A are 30°. Two light rays 
P and Q are parallel as they enter the prism. What is the angle between them after they emerge? 
  [sin
-1
 (0.8) = 53°] 
? 
Q 2.  Light is incident normally on the short face of a 30° - 60° - 90° prism. A liquid is poured on the 
hypotenuse of the prism. If the refractive index of the prism is V3, find the maximum refractive 
index of the liquid so that light is totally reflected. 
 
Q 3.  A glass vessel in the shape of a triangular prism is filled with water, and light is incident normally 
on the face XY. If the refractive indices for water and glass are 4/3 and 3/2 respectively, total 
internal reflection will occur at the glass-air surface XZ only for sin 6 greater than 
 
  A  1/2  B  2/3  C  3/4 D  8/9  E  16/27. 
Q 4.  A parallel beam of light is incident on a prism shown in figure. Through what angle should the 
mirror be rotated so that light returns back to its original path? Refractive index of prism is 1.5. 
 
Q 5.  A light ray going through a prism with the angle of prism 60°, is found to deviate at least by 30°. 
What is the range of the refractive index of the prism? 
Q 6.  A ray of light falls normally on a refracting face of a prism. Find the angle of prism if the ray just 
fails to emerge from the prism ( ? = 3/2).  
Q 7.  A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The 
ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent 
ray is perpendicular to the face through which it emerges and calculate the refractive index of the 
material of prism. 
Q 8.  A ray of light passing through a prism having refractive index 2 suffers minimum deviation. It 
is found that the angle of incidence is double the angle of refraction within the prism. What is the 
angle of prism? 
Q 9. A ray of light undergoes deviation of 30° when incident on an equilateral prism of refractive index 
2 .
 
What is the angle subtended by the ray inside the prism with the base of the prism? 
Q 10.  Light is incident at an angle i on one planar end of a transparent cylindrical rod of refractive index 
?. Find the least value of ?
 
so that the light entering the rod does not emerge from the curved 
surface of the rod irrespective of the value of i. 
 
Q 11.  The refractive index of the material of a prism of refracting angle 45° is 1.6 for a certain  
monochromatic ray. What will be the minimum angle of incidence of this ray on the prism so that 
no TIR takes place as the ray comes out of the prism. 
Solutions 
1.  
    
  ? sin i = ? sin 30° 
   = (1.6) 
1
2
??
??
??
= 0.8 
  ?   i = 53° 
  P ray deviates from its original path by an angle, ? = i - 30° = 23° 
   
  ?  Angle between two rays, ? = 2 ? 
   = 46° 
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