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Light - Reflection and Refraction : Previous Year's Questions


1 Mark Questions : - 

Q.1     Explain why a ray of light passing through the centre of curvature of a concave mirror, get s reflected along the same path.

[Delhi 2010]

Sol.      A ray of light passing through the center of curvature of a concave mirror, falls normally on the mirror and therefore retraces its path.


Q.2      What is the  nature of the image formed by a concave mirror if the magnification produced by the mirror  is + 3?

[Delhi 2010]

Sol.      Virtual.   


Q.3     Between which two points of a concave mirror should an object be placed to obtain a magnification of – 3? 

[Delhi 2010]

Sol.       Between F and C.


Q.4    "The refractive index of carbon disulphide  is 1.63." What is the meaning of this statement in relation to speed of light?

[Delhi 2010]

Sol.     The speed of light in carbon disulphide  is {1 \over {1.63}} times the speed of light in vacuum.


Q.5     The outer surface of a hollow sphere  of aluminium of radius 50 cm is to be used as a mirror. What will be the focal length of this mirror? Which type of spherical mirror will it provide?

[Delhi 2010]

Sol.      f  = 25 cm.
            Convex Mirror.


Q.6     Between which two points related to a concave mirror should an object be placed to obtain on a screen an image twice the size of the object?

[AI 2010]

Sol.       Between F and C.


Q.7    How should  a ray of light  be incident  on a rectangular glass slab so that it comes out from the opposite side of the slab without being displaced?

[Foreign 2010]

Sol.      Normally to the surface.


Q.8*    A girl was playing  with a thin beam of light from her laser torch  by directing it from different directions on a convex lens held vertically. She was surprised to see that in a particular direction the beam of light continues to move along the same direction after passing through the lens. State the reason for this observation.

[Foreign 2010]

Sol.       The beam of light must be passing through the optical centre. The reason being that the middle part of a lens  is like  a  thin glass slab, so no deviation and very slight  lateral displacement of the incident ray .


Q.9     A ray of light enters a rectangular glass slab of refractive index 1.5. It is found that the ray emerges from the opposite face of the slab without being displaced. If its speed in air is  3 \times {10^8}\,m{s^{ - 1}} then what is its speed in glass ?

[Foreign 2010]

Sol.      Speed in glass  = {{speed\,in\,air} \over {ref.index}} = {{3 \times {{10}^8}} \over {1.5}}m/s   = 2 \times {10^8}\,m/s.


Q.10     The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. What is the refractive index of the medium?

[Foreign 2010]

Sol.       ref.index = {{speed\,of\,light\,in\,vacuum} \over {speed\,of\,light\,in\,the\,medium}}  = {c \over {0.6\,c}} = 1.66


Q.11*    A ray of light LM is incident on a mirror as shown in the figure. The angle of incidence for this ray is the angle between it and the line joining two other points in the figure. Name these two points.

[CBSE 2009]

1Sol.       The angle of incidence = LMC  = angle of reflection = CMF.
              So the points are M and F.


Q.12     The following table  gives the values of refractive indices of a few media.

12Use this table to give an example of a medium pair so that light speeds up when it goes from one of these media to another.

[CBSE Sample Paper 2009]

Sol.       If light moves from diamond to any other media, it will speed up.


Q.13    A ray of light AM is incident on a spherical mirror as shown in the diagram.

2
Redraw the diagram on the answer sheet and show the path of reflected ray.

[CBSE Sample Paper 2009]

Sol.        Reflected ray is MF.
18


Q.14    Refractive index of media A, B, C and D are :

10
          In which of the four media is the speed of light                      
         (i) Maximum       (ii) Minimum

[CBSE Sample Paper 2009]

Sol.       (i) Speed of light maximum in medium A.  
             (ii) Speed of light minimum in medium D.


Q.15     Redraw the diagram given below in your answer book and show the direction of the light ray after reflection from the mirror.

[Delhi 2009 C]

3

Sol.        Reflected ray BC.


54


Q.16     Redraw the diagram given below in your answer book and show the direction of the light ray after reflection from mirror.

[Delhi 2009 C]

202

Sol.


20


Q.17     Redraw the diagram given below in your answer book and show the direction of the light ray after reflection from the mirror.

[Delhi 2009 C]

5
Sol.  

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Q.18     Redraw the diagram given below in your answer book and show the direction of the light ray after refraction from the lens.

[Delhi 2009, AI 2009 C]

6
Sol. 22


Q.19     Redraw the diagram given below in your answer book and show the direction of the light ray after refraction from the lens.

[AI 2009 C]

Q. 19
Sol.

Ans. 19


Q.20     Why does a ray of light bend when it travels from one medium into another?

[Delhi 2009]

Sol.      The ray of light bends when it travels from one medium to another because its speed is different in different media.


Q.21     If a light ray IM is incident on the surface AB as shown, identify the correct emergent ray.

[HOTS, Delhi 2008 C]

8
Sol.       Emergent ray NQ.


Q.22     The refractive indices of four media A, B, C and D are given in the following table :

11
If light travels from one medium to another, in which case the change in speed will be :
(i) Minimum , (ii) Maximum?

[Delhi 2008 C]

Sol.       (i) Between B and C.    (ii) Between A and D.


Q.23     Redraw the given diagram and show the path of the refracted ray :

[AI 2008C]

13
Sol.
24


Q.24     Redraw the given diagram and show the path of the refracted ray :

[AI 2008C]

14
Sol.
25


Q.25     “The refractive index of diamond  is 2.42”. What is the meaning of this statement in relation to speed of light?

[Delhi 2008, NCERT]

Sol.       It means that the speed of light in diamond is {1 \over {2.42}} times its speed in vacuum.


Q.26     Draw the following diagram in your answer book and show the formation of image of the object AB with the help of suitable rays.

[NCERT, AI 2008]

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Sol.       Image is A' B'


26


Q.27     Which kind of mirrors are used in the headlights of a motor- car and why?

[Foreign 2008]

Sol.       Concave mirror are used to get powerful parallel beam of light in the forward direction.


2 Marks Questions : -

Q.1     "A concave mirror of focal length ‘f ’ can form a magnified, erect as well as an inverted image of an object placed in  front of it." Justify this statement stating the position of the object with respect to the mirror in each case for obtaining these images.

[Al 2013]

Sol.      When object is placed between P and F, an erect, magnified image is formed. When the object is placed is beyond 2F, an inverted, magnified image is formed between F & 2F.


Q.2     List four properties of the image formed by a plane mirror.

[Foreign 2013, Delhi 2012, Al 2011]

Sol.      Properties of image formed by a plane mirror-
           
(i) Virtual
            (ii) Erect
            (iii) Same size as object
            (iv) As far behind the mirror as the object is in front of it.


Q.3     List four properties of the image formed by a convex mirror.

[Delhi 2012]

Sol.       Properties of image formed by convex mirror-
             (i) Virtual
             (ii) Erect
             (iii) Diminished
             (iv) Image lies between P and F.


Q.4     List four properties of the image formed by a concave mirror, when object is placed between focus and pole of the mirror.

[Delhi 2012]

Sol.       Properties of image formed by concave mirror when object lies between P and F.
             (i) Virtual
             (ii) Erect
            (iii) Magnified
            (iv) Formed behind the mirror.


Q.5     To construct a ray diagram, we use two light rays which are so chosen that it is easy to know their directions after refraction from the lens. List these two rays and state the path of these rays after refraction. Use these two rays to locate the image of an object placed between ‘F’ and ‘2F’ of a convex lens.

[Foreign 2012]

Sol.       (i) Any ray parallel to the principal axis  after refraction will pass through the focus. 
             (ii) Any ray passing  through the focus, after refraction will go parallel to the principal axis.

272


Q.6    State the two laws of reflection of light.

[Delhi 2011]

Sol.      Laws of reflection
            I Law : The incident-ray, the reflected ray and the normal (at the point of incidence) all lie in the same plane.
            II Law : The angle of reflection is equal to the angle of incidence.

30


Q.7     Define and show on a diagram, the following terms relating to a concave mirror:
          
(i) Aperture     (ii) Radius of curvature

[Foreign 2011]

Sol.        (i) The diameter of the concave mirror is called its aperture.
              (ii) The radius of the sphere of which the concave mirror is a part of is called its radius of curvature. R = CP 

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             PC is radius of curvatures, ab is aperture.


Q.8     Define the focus of a concave mirror. If the radius of curvature of a convex mirror is 30 cm, what would be its focal length?

[Foreign 2011]

Sol.       When rays parallel and close to the principal axis of a concave mirror meet at a point , this point is called "principal focus".
              f = {{ R} \over 2} = {{ - 30} \over 2} = - 15 cm.


Q.9     Distinguish between a real and a virtual image of an object. What type of image is formed
           (i) by a plane mirror,      (ii) on a cinema screen 

[Foreign 2011]

Sol.       (i) Virtual image is formed when diverging rays reach the eye. They appear to be coming from a point, although, there is no actual intersection of rays. e.g. image formed by plane mirror.
             (ii) Real image is formed by the actual intersection of rays. It can be formed on a screen e.g. image formed on a cinema screen.


Q.10     In an experiment with a rectangular glass slab, a student observed that a ray of light incident at an angle of 55º with the normal on one face of the slab, after refraction strikes the opposite face of the slab before emerging out into air making an angle of 40° with the normal. Draw a labelled diagram to show the path of this ray. What value would you assign to the angle of refraction and angle of emergence ?

[HOTS. AI 2010]

Sol.       Angle of refraction  = 40^\circ
                 Angle of emergence  = 55^\circ

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Q.11    What is the minimum number of rays required for locating the image formed by a concave mirror for an object. Draw a ray diagram to show the formation of a virtual image by a concave mirror.

[Delhi 2009]

Sol.  Two rays.

10


Q.12     The refractive index of water is 1.33 and the speed of light in air is 3 \times {10^8}m{s^{ - 1}}. Calculate the speed of light in water.

[Foreign 2009]

Sol.        Speed of light in water  = {{speed\,of\,light\,in\,vacuum} \over {ref.index}}
                = {{3 \times {{10}^8}} \over {1.33}}
                = 2.25 \times {10^8}m/s


Q.13    The refractive index of glass is 1.50 and the speed of light in air is 3 \times {10^8}m{s^{ - 1}}.  Calculate the speed of light in glass.

[Foreign 2009]

Sol.        Speed of light in glass  = {{speed\,of\,light\,in\,vacuum} \over {refractive\,index}}
                = {{3 \times {{10}^8}} \over {1.5}}m/s.
                = 2 \times {10^8}m/s.


Q.14    Explain with the help of a diagram, why a pencil partly immersed in water appears to be bent at the water surface.

[Delhi 2008]

Sol.        When rays of light move from water to air they bend away from the normal and appear to be coming from a point higher than its actual position. So, the pencil appears to be bent.

Q. 14


Q.15    Draw ray diagrams to represent the nature, position and relative size of the image formed by a convex lens for the object placed: 
              (a) at 2{F_1}
              (b) between {F_1} and the optical centre O of lens.

[Al 2008]

Sol.        (a)l7             (b)l8


Q.16     A ray of light, incident obliquely on a face of a rectangular glass slab placed in air, emerges from the opposite face parallel to the incident ray. State two factors on which the lateral displacement of the emergent ray depends.

[Foreign 2008]

Sol.         (i) Angle of incidence     (ii) Thickness of glass slab.


Q.17    Draw a ray diagram to show the (i) position and (ii) nature of the image formed when an object is placed between focus F and pole P of a concave mirror.

[Delhi 2006]

Sol.10
                 (i) Image formed behind the mirror                                      
            (ii) Image is virtual, erect and magnified.


Q.18     An object is placed at a distance of 12 cm in front of a concave mirror. It forms a real image four times larger than the object. Calculate the distance of the image from the mirror.

[Al 2006]

Sol.         u = – 12 cm ; m = – 4.
               m = {{ - v} \over u}
               Therefore,  v = – (– 4) × (–12) = – 48 cm.
               Image is formed 48 cm in front of mirror.


3 Marks Questions : -

Q.1     An object placed on a meter scale at 8 cm mark was focused on a white screen placed at 92 cm mark, using a converging lens placed on the scale at 50 cm mark.
          
(a) Find the focal length of converging lens.
        
(b) Find the position of image formed if the object is shifted towards the lens at a position of 29.0 cm.
         
(c) State the nature of image formed if the object is further shifted towards the lens.

[Al 2013]

Sol.        u = 8 – 50 = – 42 cm.
               v = 92 – 50 = 42 cm.
              (a) {1 \over f} = {1 \over v}\, - \,{1 \over u}\,
                     = {1 \over {42}}\, + \,{1 \over {42}} = {2 \over {42}}
                    Therefore,  f  = 21 cm.
              (b) u = 29 – 50 =  – 21 cm.
                    Since object is at focus, image is formed at infinity.
              (c) Virtual image, erect and magnified.


Q.2    State the type of mirror preferred as (i) rear view mirror in vehicles, (ii) shaving mirror. Justify your answer giving two reasons in each case.

[Delhi 2013, 2012]

Sol.        (i) Convex mirror to gives wider field of view and it produces erect, diminished image.
              (ii) Concave mirror is used. It gives erect magnified image, if object is at a distance lesser than its focal length.


Q.3    The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.

[Delhi 2012]

Sol.

u = – 45 cm ; v = 90 cm.
{1 \over f} = {1 \over v} - {1 \over u}
 = {1 \over {90}} - \left( {{{ - 1} \over {45}}} \right) = {1 \over {30}}
f = + 30.
Convex lens of focal length 30 cm.
{{h'} \over h} = {v \over u}
h' = {{2 \times 90} \over { - 45}} =- \,4cm.
Height of image = 4 cm.


Q.4     A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.

[Foreign 2013, Al 2012]

Sol.  

h = 4 cm.
f = 24 cm.
u = – 16 cm.
{1 \over f} = {1 \over v} - {1 \over u}
\,{1 \over v} = {1 \over {24}} + \left( {{{ - 1} \over {16}}} \right)
 = {{ - 1} \over {48}}
Therefore, v = – 48 cm.
h' = h \times {v \over u}
 = 4 \times {{ - 48} \over { - 16}} = 12 cm.
Image is formed at 48 cm. in front of lens. It is a virtual, erect image of size 12 cm.


Q.5     Name the type of mirror used in the following situations:
          
(i) Headlights of a car
          
(ii) Rear-view mirror of vehicles
          
(iii) Solar furnace

              Support your answer with reason.

[Foreign 2013]

Sol.       (a) Concave mirror - The bulb placed at focus throws parallel beam of light after reflection.
              (b) Convex mirror - To get wider field of view and erect images.
              (c) Concave mirror - To converge the rays of sun at its focus.


Q.6    A 6 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 15 cm. The distance of the object from the lens is 10 cm. Find the position, size and nature of the image formed, using the lens formula.

[Delhi 2013, Foreign 2012]

Sol.  

h = 6 cm.
f = 15 cm.
u = – 10 cm.
{1 \over v} = {1 \over f} + {1 \over u}
 = {1 \over {15}} - {1 \over {10}} = {{ - 1} \over {30}}
Therefore,  v = – 30 cm.
{\rm{h' = h}} \times \left( {{{\rm{v}} \over {\rm{u}}}} \right)
6 \times {{ - 30} \over { - 10}} = 18 cm.
Image is formed 30 cm on the same side of lens. It is erect, magnified, virtual having height 18 cm..


Q.7     Draw a ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed. at the centre of curvature of the mirror.

[Delhi 2011]

Sol.1


Q.8    Define, 'refractive index of a transparent medium.' What is its unit? Which has a higher refractive index, glass or  water?

[Delhi 2011]

Sol.        The refractive index of a transparant medium is the ratio of the speed of light in vacuum to that in the medium-
               n = {c \over v}
               It has no unit.

                   Glass has higher refractive index.


Q.9    A ray of light travelling in air enters obliquely into water. Does  the  light  ray  bend  towards  or  away  from  the normal ? Why? Draw a ray diagram to show the refraction of light in this situation.

[Delhi 2011]

Sol.

51

             It bends towards normal, as water is optically denser than air.


Q.10
         (a) "The refractive index of diamond is 2.42”. What is the meaning of this statement?
         (b) Name a liquid whose mass density is less than that of water but it is optically denser than water.

[Delhi 2011]

Sol.       (a)  It means that the speed of light in diamond is {1 \over {2.42}} times its speed in vacuum.
             (b) Kerosene.


Q.11     An object is placed between infinity and the pole of a convex mirror. Draw a ray diagram and also state the position, the relative size and the nature of the image formed.

[Al 2011]

Sol. l10
              The image is formed between P and F. It is diminished, virtual and erect.


Q.12    What is the principle of reversibility of light? Show that the incident ray of light is parallel to the emergent ray of light when light falls obliquely on a side of a rectangular glass slab.

[AI 2011]

Sol.       The  principal of reversibility states that if the final path of ray of light after reflection or refraction is reversed, the ray retraces its entire path.

Q.12

 


Q.13    With the help of a ray diagram explain why a convex minor is preferred for rear view mirrors in motor cars.

[Foreign 2011]

Sol.  Greater field of view and image is always erect.

33


Q.14    What  is  understood  by  lateral  displacement  of  light? Illustrate it with the help of a diagram. List any two factors on which the lateral displacement of a particular substance depends.

[Foreign 2011]

Sol.       When a ray of light passes obliquely through a glass slab, the emergent ray is found to have shifted parallel to the direction of incident ray. This is called lateral  shift.
light1
               Lateral shift depends on
              (i) Angle of incidence
              (ii) Thickness of glass slab.


Q.15    Draw the ray diagram and also state the position, relative size and nature of the image formed by a concave mirror when the object is placed between its centre of curvature C and focus F.

[Foreign 2011]

Sol. l11

Image is formed beyond C ,
Real
Magnified
Inverted.


Q.16    At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side. What will be the magnification produced in this case?

[Delhi 2010]

Sol.  

f = 18 cm.
v = 24 cm.
{1 \over u} = {1 \over v} - {1 \over f}
 = {1 \over {24}} - {1 \over 18}
 = {{ - 1} \over {72}}
Therefore,  u = – 72 cm.
m = {v \over u}\, = \,{{ - 24} \over {72}}\, = \,{{ - 1} \over 3}
Magnified  = {{ - 1} \over 3}.


Q.17    How far should an object be placed from a convex lens of  focal length 20 cm to obtain its image at a distance of 30 cm from the lens? What will be the height of the image if the object is 6 cm tall?

[AI 2010]

Sol.  

f = 20 cm.
v = 30 cm.
{1 \over u} = {1 \over v} - {1 \over f}
= {1 \over {30}}\, - \,{1 \over {20}}
= {{ - 1} \over {60}}
Therefore,  u = – 60 cm.
{{h'} \over h} = {v \over u}
h' = 6 \times {{30} \over { - 60}} =- 3cm
The object should be placed 60 cm from the lens. The image will be real, inverted and of size 3 cm.


Q.18    The image of an object placed at 60 cm in front of a lens is obtained on a screen at a distance of 120 cm from it. Find the focal length of the lens. What would be the height of the image if the object is 5 cm high?

[Foreign 2010]

Sol.  

u = – 60
v = 120 cm.
{1 \over f} = {1 \over v}\, - \,{1 \over u}
= {1 \over {120}}\, + \,{1 \over {60}} = \,{3 \over {120}} = {1 \over {40}}
Therefore,  f = 40 cm.
h' = h \times {v \over u}
 = 5 \times {{120} \over { - 60}} =- 10\,cm.
The focal length is 40 cm, and height of image is 10 cm.


Q.19    (a) Two lenses have power of
(i) + 2 D
(ii) – 4 D.
What is the nature and focal length of each lens?

(b) An object is kept at a distance of 100 cm from each of the above lenses.
Calculate the (i) image distance (ii) magnification in each of the two cases.

[CBSE Sample Paper 2009]

Sol.       (a)  (i) Convex lens.
                         f = {1 \over 2} = 0.50 m.
                     (ii) Concave lens.
                           f = {{ - 1} \over 4} = -0.25 m.
              (b)  Case I. convex lens.
                     f = 50 cm ; u = – 100 m.
                     {1 \over v} = {1 \over f}\, + \,{1 \over {u\,}}
                      = {1 \over {50}}\, - \,{1 \over {100}} = {1 \over {100}}
                      v = 100 cm.
                      m = {v \over u} = {{100} \over { - 100}} = \, - 1
                     Case II.
                    
f = – 25 cm ; u = -100 cm.
                     {1 \over v} = {1 \over u}\, + \,{1 \over f}
                     = {{ - 1} \over {100}}\, - \,{1 \over {25}}
                     = {{ - 1} \over {100}}\, - \,{4 \over {100}}
                     = {{ - 1} \over {20}}
                     Therefore,  v = – 20 cm.
                     m = {v \over u} = {{ - 20} \over { - 100}} = 0.2


Q.20    For which position of the object does a convex lens form a virtual and erect image? Explain with the help of a ray diagram.

[Al 2009]

Sol.        In a convex lens, a virtual, erect image is formed when object distance is less than the focal length.
l8


Q.21    An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image fanned? Draw a ray diagram to show the formation of the image in this case.

[Delhi 2008 C]

Sol.  

h = 2 cm.
u = – 30 cm.
f = – 15 cm.
{1 \over v} = {1 \over f}\, - \,{1 \over u}
 = {{ - 1} \over {15}}\, + \,{1 \over {30}}
 = {{ - 1} \over {30}}
Therefore,  v = – 30 cm.
m = {{ - v} \over u}\, = \, - \,\left( {{{ - 30} \over { - 30}}} \right) = \, - 1.
The screen should be placed as 30 cm from the mirror. The image will be real, inverted and same size as the object.
l3


Q.22    An object 2 cm high is placed at a distance of 64 cm from a white screen. On placing a convex lens at a distance of 32 cm from the object it is found that a distinct image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.

[Delhi 2008 C]

Sol.  

h = 2 cm.
u + v = 64 cm (Total distance)
u = – 32 cm.
v = 32 cm.
{1 \over f} = {1 \over v}\, - \,{1 \over u}\,
 = {1 \over {32}}\, + \,{1 \over {32}}\, = {1 \over {16}}
f = 16 cm.
m = {v \over u} = {{ - 32} \over {32}} = - 1.
h' = -2 cm.l5


Q.23    A  convex  lens  has  a  focal  length  of  10  cm.  At  what distance from the lens should the object be placed so that it forms a real and inverted image 20 cm away from the lens? What would be the size of the image formed if the object is 2 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.

[AI 2008 C]

Sol.  

f = 10 cm.
v = 20 cm.
f = 2 cm.
{1 \over u} = {1 \over v} - {1 \over {f\,}}
 = {1 \over {20}} - {1 \over {10}}\, = \,{{ - 1} \over {20}}
Therefore,  u = – 20 cm.
h' = h \times {v \over u}
 = 2 \times {{20} \over { - 20}} = - 2\,cm.
l5


Q.24    A  concave  lens  has  focal  length  of  20  cm.  At  what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

[Delhi 2007]

Sol.  

f = – 20 cm.
h = 5 cm.
v = –15 cm.
{1 \over v}\, - \,{1 \over u} = \,{1 \over f}
{1 \over u} = {1 \over v}\, - \,{1 \over f}
 = {{ - 1} \over {15}}\, - \left( {{{ - 1} \over {20}}} \right)
 = {{ - 1} \over {60}}
Therefore,  u = – 60 cm.
{{h'} \over h} = {v \over u}
h' = 5 \times {{ - 15} \over { - 60}} = {5 \over 4}cm
Object must be placed 60 cm. Size of image is 1.25 cm.


Q.25    An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.

[AI 2007]

Sol.  

h = 50 cm.
h' = - 20 cm.
v = 10 cm.
{{h'} \over h} = {v \over u}
{{ - 20} \over {50}} = {{10} \over u}
Therefore u = – 25 cm.
{1 \over f} = {1 \over v} - {1 \over u}
= {1 \over {10}} - \left( {{{ - 1} \over {25}}} \right) = {{35} \over {250}}
f = {{50} \over 7} = 7.14\,cm.


Q.26    A 5.0 cm  tall  object  is  placed  perpendicular  to  the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. By calculation determine (i) the position and (ii) the size of the image formed.

[Delhi. AI 2006]

Sol.

(i) h = 5 cm.
f = 20 cm.
u = – 30 cm.
{1 \over v} = {1 \over f} + {1 \over u}
= {1 \over {20}} - {1 \over {30}} = {1 \over {60}}
Therefore, v = 60 cm.
(ii)  h' = h \times {v \over u}
 = 5 \times {{60} \over { - 30}} = - 10\,cm.


5 Marks Questions : -

Q.1    (a) To construct a ray diagram, we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror.

[Al 2012]

(b) Draw a ray diagram to show the formation of image of an object placed between the pole and principal focus of a concave mirror. How will the nature and size of the image formed change, if the mirror is replaced by converging lens of same focal length ?

[AI 2013]

Sol.     (a) (i) Any ray parallel to the principal axis, after reflection from a concave mirror passes through the focus.
                 (ii) Any ray passing through the focus will go parallel to the principal axis after reflection.

41

            (b) If the mirror is replaced by a convex lens of same focal length the image will be formed on same size as object. The nature and size of image will not change.
light2


Q.2     (a) With the help of a ray diagram, state the meaning of refraction of light. State Snell’s law of refraction of light and also express it mathematically.
(b) The refractive index of water with respect to vacuum is 4/3 and refractive index of vacuum with respect to glass is 2/3. If the speed of light in glass is 2 \times {10^8}\,m{s^{ - 1}} , find the speed of light in (i) vacuum, (ii) water.

[Foreign 2013]

Sol.       (a) Snell's law states that the ratio of {{\sin \,i} \over {\sin \,r}} is a constant for a given pair of medium and is called the refractive index of the second medium w.r.t the first.
             It is denoted by n_{21}
             n_{21} = {{\sin \,i} \over {\sin \,r}}
35

(b) n_{wv} = {4 \over 3}
n_{vg}= {2 \over 3}
(i) n_{vg}= {2 \over 3} = {{speed\,(glass)} \over {speed\,in\,vacuum}}
speed\,in\,vacuum\, = {{speed\,of\,light\,in\,glass} \over {2/3}}
= {{2 \times 10^8\times 3} \over 2} = 3 \times 10^8 \,m/s
n_{wv}= {4 \over 3}
 = {{speed\,of\,light\,in\,vacuum} \over {speed\,of\,light\,in\,water}}
Therefore,  speed of light in water = {{3 \times 10^8 } \over {{4 \over 3}}}
= {9 \over 4} \times 10^8 \,m/s.
= 2.25 \times 10^{8\,} \,m/s.


Q.3    List the sign conventions for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three times magnified real image of an object placed 16 cm in front of it.

[Delhi 2012]

Sol.        Sign conventions for spherical mirrors.

(a) (i) All measurements are taken from the pole of the mirror.
(ii) Measurements in the direction of incident light are taken as positive.
Measurement against the incident light are taken as negative.
(iii) Distances measured above and perpendicular to the principal axis are taken as positive and below it are taken as negative
(b)  Q.3
OP is – ve i.e. u is -ve.
IP is – ve i.e. v is -ve.
m = – 3. Image is inverted.
m = – 3 = {{ - v} \over u}
Therefore,  v = 3u.
u = – 16 cm ; v = – 48 cm.
{1 \over f} = {1 \over v} + {1 \over u}
{{ - 1} \over {48}} - {1 \over {16}}
 = {{ - 4} \over {48}} = {{ - 1} \over 2}
Therefore ,  f = – 12 cm.


Q.4 State  the  law  of  refraction  of  light  that  defines the refractive index of a medium with respect to the other. Express it mathematically. How is refractive index of any medium 'A' with respect to a medium 'B' related to the speed of propagation of light in two media A and B? State the name of this constant when one medium is vacuum or air. The refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 respectively. If the speed of light in glass is 2 \times {10^8}\,m/s  find the speed of light in (i) vacuum, (ii) water.

[Delhi 2013, 2012]

Sol.

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given color and for  the given pair of media .This law is known as Snell's Law .
{{Sin\,i} \over {Sin\,r}} = constant = refractive index of second medium with respect to the first medium
where i is the angle of incidence
where i is the angle of refraction
{n_{AB}} = {{{v_B}} \over {{v_A}}}
Refractive index of A w.r.t.
{v_B} is velocity of light in medium B.
{v_A } is velocity of light in medium A.
When one medium is vacuum, it is called absolute refractive index.
n_{wv} = {4 \over 3}

n_{vg} = {2 \over 3}
(i) n_{vg}= {2 \over 3} = {{speed\,(glass)} \over {speed\,in\,vacuum}}

speed\,in\,vacuum\, = {{speed\,of\,light\,in\,glass} \over {2/3}}
= {{2 \times 10^8\times 3} \over 2} = 3 \times 10^8 \,m/s
n_{wv}= {4 \over 3}
 = {{speed\,of\,light\,in\,vacuum} \over {speed\,of\,light\,in\,water}}
Therefore,  speed of light in water = {{3 \times 10^8 } \over {{4 \over 3}}}
= {9 \over 4} \times 10^8 \,m/s.
= 2.25 \times 10^{8\,} \,m/s.


Q.5     List the new Cartesian sign convention for reflection of light  by  spherical  mirrors.  Draw a diagram  and  apply these  conventions  for  calculating  the  focal  length  and nature  of  a spherical  mirror  which  forms  a  1/3  times magnified virtual image of an object placed 18 cm in front of it.

[Al 2012]

Sol. 

(i) All measurements are taken from the pole of the mirror.
(ii) Measurements in the direction of incident light are taken as positive.Measurement against the incident are taken as negative.
(iii) Distances measured above and perpendicular to the principal axis are taken as positive and below
it are taken as negative.


Q. 5

 
 m = {{{\rm{ + 1}}} \over {\rm{3}}} as image is erect

u = – 18 cm.
 m = {{\rm{1}} \over {\rm{3}}} = {{ - v} \over u}
\,v = {u \over 3} = {{ - 18} \over { - 3}} = 6\,cm
{1 \over f} = {1 \over v} + {1 \over u}
 = {1 \over 6} - {1 \over {18}} = {2 \over {18}}
Therefore,  f = + 9 cm.


Q.6 List  the  sign  conventions  that  are  followed  in  case of refraction of  light through spherical  lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms three times magnified  real image of an object placed 16 cm from the lens.

[Foreign 2012]

Sol.

new1u = – 16
 m = {{ - 1} \over 3} Image as inverted.
 m = {{ - 1} \over 3} = {v \over u}
\,\,v = {{ - 4} \over 3}
 = {{ - \left( { - 16} \right)} \over 3} = {{16} \over 3}
{1 \over f} = {1 \over v} - {1 \over u}
 = {3 \over {16}} - \left( {{{ - 1} \over {10}}} \right)
 = {1 \over 4}
Therefore,  f = +4 cm.


Q.7 State the laws of refraction of light. Write an expression to relate absolute refractive index of a medium with speed of light in vacuum.
The refractive index of a medium 'x' with respect to 'y' is 2/3 and the refractive index of medium 'y' with respect to ‘z’ is 4/3. Calculate the refractive index of medium ‘z’ with respect of 'x'.

[Foreign 2012]

Sol.        Laws of refraction of light

(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at a point of incidence ,all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is known as Snell's law.
{{\sin i} \over {\sin r}}= Constant = Refraction index of second medium w.r.t. the first medium
Where i is angle of incidence
r is angle of refraction
{n_{xy}} = {2 \over 3}
{n_{yz}} = {4 \over 3}
Since, {n_{xy}} \times {n_{yz}} \times {n_{zx}}
Therefore, \,{n_{zx}} = {1 \over {{n_{xy}} \times {n_y}}}
 = {1 \over {{2 \over 3} \times {4 \over 3}}} = {9 \over 8}


Q.8    (a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it?
(b) Name the point on the lens through which a ray of light passes undeviated.

(c) An object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the position (ii) the magnification and (iii) the nature of the image formed.

[Delhi 2011]

Sol.  

(a) Convex lens.
(b) Optical centre
(c) f = 20 cm
u = – 30 cm.
{1 \over v} = {1 \over f} + {1 \over u}
 = {1 \over {20}} - {1 \over {30}} = {1 \over {60}}
(i) v = 60 cm.
(ii)  m = {v \over u} = {{60} \over { - 30}} = - 2.
(iii) Image is real, inverted, magnified.


Q.9    (a) What is meant by `power of a lens'?
(b) State and define the S.I. unit of power of a lens.
(c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination.

[A1 2011]

Sol.  

(a) Power of  a lens is the degree of its convergence or divergence . It is the reciprocal of its focal length in metres. Its unit is dioptres.
(b) S.I. unit of power is dioptre .One dioptre is the power of lens whose focal length is 1 m.
(c) {f_1} =+ 25 cm\,\,;\,\,{f_2} =- 10\,cm.
{1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}}
 = {1 \over {25}} - {1 \over {10}} = {{ - 3} \over {50}}
 P = {1 \over {f(m)}} = {{ - 3 \times 100} \over {50}} =- 6\,D


Q.10    (a) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens.
(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate
(i) the distance of the object from the lens
(ii) the magnification for the image formed
(iii) the nature of the image formed

[Al 2011]

Sol.  

(a)

Q. 10


(b) f = – 15 cm ; v = –10 cm.
(i) {1 \over u} = {1 \over v} - {1 \over f}
= {{ - 1} \over {10}} - \left( {{{ - 1} \over {15}}} \right)
 = {{ - 1} \over {10}} + {1 \over {15}} = {{ - 1} \over {30}}
Therefore,  u = – 30 cm.
(ii)  m\, = {{ + v} \over u}
 = {{( - 10)} \over {( - 30)}} = {1 \over 3}
(iii) Image is virtual, erect and diminished.


Q.11    (a) Under what condition will a glass lens placed in a transparent liquid become invisible?
(b) Describe and illustrate with a diagram, how we should arrange two converging lenses so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens.
(c) An object is placed at a distance of 3 cm from a concave lens of focal length 12 cm. Find the (i) position and (ii) nature of the image formed.

[Foreign 2011]

Sol.     (a) When the refractive index of glass and liquid are same, the glass lens will become invisible in the liquid.
          (b) When both the converging lenses are placed at a distance of f_1 + f_2 where f_1 is the focal length of lens 1 and f_2 is the focal length of lens 2.

Q. 11

               

                 (c) u = – 3 cm.
                  f = – 12 cm.
                 (i) {1 \over v} = {1 \over f} + {1 \over u}
                       = {{ - 1} \over {12}}\, - \,{1 \over 3} = {{ - 5} \over {12}}
                      \,v = {{ - 12} \over 5}\,cm = - 2.4\,cm
                 (ii)  m = {{ + v} \over u} = {{ - 2.4} \over { - 3}} = 0.8
              Image is virtual, erect, diminished.


Q.12    (a) With the help of a ray diagram explain why a concave lens diverges the rays of a parallel beam of light.
(b) A 2.0 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15 cm. At what distance from the lens, should the object be placed so that it forms an image 10 cm from the lens? Also find the nature and the size of image formed.

[Foreign 2011]

Sol. (a)

Q. 12

When a ray of light is incident on the surface of the concave lens, it bends towards the normal.
At the second interface it bends away from the normal due to its curvature AB bends away from its straight line path and becomes BC and then it bends away to become CD. In both cases it moves away from its incident path.
Thus the ray becomes diverging
(b) h = 2 cm.
f = – 15 cm.
v = – 10 cm.
{1 \over u} = {1 \over v} - {1 \over f}
 = {{ - 1} \over {10}} - \left( {{{ - 1} \over {15}}} \right) = {{ - 1} \over {30}}
Therefore,  u = – 30 cm.
 m = {v \over u} = {{ - 10} \over { - 30}} = {{ - 1} \over 3}
h' = m \times h
 = {1 \over 3} \times 2 = {2 \over 3}\,cm.
The object should be placed at a distance of 30 cm. The image is virtual, erect, diminished of size 0.67 cm.


Q.13    Draw ray diagrams to show the formation of a three times magnified (i) real image (ii) virtual image of an object kept in front of a converging lens. Mark the positions of object, F, 2 F, O and position of image clearly in the diagram. 
An object of size 5 cm is kept at a distance of 25 cm from the Optical centre of a converging lens of focal length 10 cm. Calculate the distance of the image from the lens and size of the image.

[CBSE Sample Paper 2009]

Sol.     (i) 46
(ii)

Q. 13

h = 5cm.
u = – 25cm.
f = 10 cm.
{1 \over v} = {1 \over f} + {1 \over u}
 = {1 \over {10}} - {1 \over {25}} = {3 \over {50}}
\,v = {{50} \over {30}} = 16.6\,cm.
{{h'} \over h} = {v \over u}
h' = 5 \times {{{{50} \over 3}} \over { - 25}}
 = {{ - 10} \over 3}\,cm\, = \, - 3.3

 


Q.14    (a) It is desired to obtain an erect image of an object, using a concave mirror of focal length 20 cm.
(i) What should be the range of distance of the object from the mirror?
(ii) Will the image be bigger or smaller than the object?
(iii) Draw a ray diagram to show the image formation in this case.
(b) One half a convex lens of focal length 20 cm is covered with a black paper.
(i) Will the lens produce a complete image of the object?
(ii) Show the formation of image of an object placed at 2F, of such covered lens with the help of a ray diagram.
(iii) How will. the intensity of the image formed by half-covered lens compare with non-covered lens?

[HOTS, Foreign 2008]

Sol.     (a) (i) The range should be 0 to 20 cm.
                 (ii) Image will be bigger
                 (iii)10
              (b) (i) Yes.
                    (ii) 48
                      (iii) The intensity of image will be halved, when half the lens is covered.


Q.15    Draw the ray diagram in each case to show the position and nature of the image formed when the  object is placed :    (i) At the centre of curvature of a concave minor
(ii) Between the pole P and focus F of a concave mirror
(iii) In front of a convex mirror
(iv) At 2F of a convex lens
(v) In front of a concave lens

[Delhi 2007]

Sol.       (i) l3
               (ii)10
                  (iii) l4
                   (iv) l5
                    (v) l6



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