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Light - Reflection and Refraction : S.Chand


Q.1     State the laws of reflection of light at a plane surface (like a plane mirror), and mark the angles of incidence and reflection on a diagram.
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 angel of reflection is equal to the angle of incidence.

27


Q.2    Define the following terms used in the reflection of light by drawing a labelled diagram :
(a) Incident ray
(b) Normal

(c) Reflected ray
(d) Angle of incidence

(e) Angle of reflection
Sol. 28

(a) AO is incident ray
(b) ON is normal

(c) OB is reflected ray
(d) \angle AON is angle of incidence

(e) \angle NOB is angle of reflection


Q.3    What happens when a ray of light falls ‘normally’ on the surface of a mirror ?
Sol.      When a ray fall ‘normally’ on the surface of a mirror it is reflected back.


Q.4    A ray of light is incident on a plane mirror at an angle of 30°. What is the angle of reflection?
Sol.     \angle r = 30^\circ


Q.5    A ray of light is incident normally on a plane mirror. What will be the :
(a) angle of incidence ?                                      
(b) angle of reflection ?

Sol.     (a) \angle i = 0                                        
           (b) \angle r = 0


Q.6    If an object is placed at a distance of 10 cm in front of a plane mirror, how far would it be from its image ?
Sol.      20 cm


Q.7    Draw a labelled diagram showing how a plane mirror forms as image of a point source of light in front of it.
Sol. 
Q. 7 schand

The rays appear to be coming from A’B’
A’B’ is the virtual image.


Q.8    What is the difference between a real image and a virtual image ? Give one example of each type of image.
Sol.        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.
               Virtual image is formed when diverging rays reach the eye. The appear to be coming from a point, although, there is no actual intersection of rays. E.g. image formed by plane mirror.


Q.9    What type of images is formed :
(a) in a plane mirror ?
(b) on a cinema screen ?

Sol.     (a) Virtual                                                          
           (b) real


Q.10   State four characteristics of the image formed in a plane mirror.
Sol.     Virtual, erect, same size as object, as far behind the mirror and the object is in front of it.


Q.11   What kind of mirror is required for obtaining a virtual image of the same size as the object?
Sol.      Plain mirror


Q. 12    The letter F is placed in front of a plane mirror :
(a) How would its image look like when seen in a mirror ?
(b) What is the name of the phenomenon involved ?
Sol.       (a)
                    

                (b) Lateral inversion


Q.13   Why does a parting in our hair on the left appear to be a parting on the right when seen in a mirror ?
Sol.        Due to lateral inversion of the image.


Q.14    Name the phenomenon responsible for the following effect : When we sit in front of a plane mirror and write with our right hand, it appears in the mirror that we are writing with the left hand.
Sol.        Due to lateral inversion of the image.


Q.15     What is lateral inversion ? Explain by giving a suitable example.
Sol.        The sideways inversion of an image formed by a plane mirror is called lateral inversion. e.g. letter R appears as
             


Q.16    What is the name of the phenomenon in which the right side of an object appears to be left side of the image in a plane mirror ?
Sol.        Due to lateral inversion of the image.


Q.17    Define the principal focus of a concave mirror.
Sol.      When rays parallel and close to the principal axis of a concave mirror meet at a point, this point is called principal focus.


Q.18    Draw a labelled diagram to show : (i) centre of curvature, (ii) principal axis, (iii) principal focus, and (iv) pole, of a concave mirror.
Sol.     1

(i) C      (ii) PC         (iii) F          (iv) P


Q.19    Define focal length and radius of curvature for a spherical mirror.
Sol.        Focal length is the distance between pole and focus of the mirror.
                 Radius of curvature is the radius of the sphere of which the curved mirror is a part of.


Q.20    Explain with a suitable diagram, how a concave mirror converges a parallel beam of light rays.
Sol. GRayscale


Q.21    Explain with a suitable diagram, how a convex mirror diverges a parallel beam of light rays. Mark the focus.
Sol.   4


Q.22    What is a spherical mirror ? Distinguish between a concave mirror and a convex mirror.
Sol.        A spherical mirror is a part of a hollow sphere.
                 When the inner part is reflecting surface, it is called concave mirror and when outer part is reflecting surface it is called converse mirror.
                 In concave mirror real, inverted images are formed or virtual enlarged image.
                 The image formed by convex mirror is always virtual, erect and diminished.


Q.23    (a) Define the principal focus of a convex mirror.
              (b) Out of convex mirror and concave mirror, whose focus is situated behind the mirror ?
Sol.      When rays parallel to the principal axis are incident on a convex mirror, they are reflected and appear to diverge from a point. This point is called the focus.


Q.24    Name the spherical mirror which has :
                 (a) virtual principal focus                                     
          
(b) real principal focus

Sol.       (a) Convex mirror                                                    
             (b) Concave mirror


Q.25   What is the relation between the focal length and radius of curvature of a spherical mirror (concave mirror or convex mirror) ?
Sol.       Focal length ={1 \over 2} of radius of curvature i.e. f={R \over 2}


Q.26    Calculate the focal length of a mirror whose radius of curvature is 25 cm.
Sol.        f = {R \over 2}
                           = {{25} \over 2}
                  f = 12.5 cm


Q.27    If the focal length of a convex mirror is 25 cm, what is its radius of curvature ?
Sol.        R = 2f = 2 × 25 cm = 50 cm


Q.28    Find the focal length of a concave mirror whose radius of curvature is 32 cm.
Sol.       f = {R \over 2}
                           = {{32} \over 2} cm = 16 cm


Q.29    For what position of an object, a concave mirror forms a real image equal in size to the object ?
Sol.       Object is at 2F (or C)


Q.30    For what position of an object, a concave mirror forms an enlarged virtual image ?
Sol.       Object is between pole and focus.


Q.31  Where should an object be placed in front of the concave mirror so as to obtain its magnified erect image ?
Sol.       Object is between pole and focus.


Q.32    For which positions of the object does a concave mirror produce an inverted, magnified and real image ?
Sol.       When object is between F & C.


Q.33    Describe with the help of a diagram, the nature, size and position of the image formed when an object is placed between the pole and focus of a concave mirror.
Sol. 

Q. 33 schand


Q.34   An object is placed in front of concave mirror between its focus and centre of curvature. Show with the help of a ray diagram, the position, size and nature of the image formed.
Sol. 
Q.34 schand


Q.35   With the help of a ray diagram, determine the position, nature and size of the image formed of an object placed at the centre of curvature of a concave mirror.
Sol. 35


Q.36    Describe with the help of a diagram, the nature, size and position of the image formed when an object is placed beyond the centre of curvature of a concave mirror.
Sol.7


Q.37    If an object is placed at a distance of 8 cm from a concave mirror of focal length 10 cm, discuss the nature of the image formed by drawing the ray diagram.
Sol. 

31

Image is virtual, magnified and formed behind the mirror.


Q.38    Give at least three uses of a concave mirror.
Sol.       Uses of concave mirrors
                (i) Used as reflection in torches, vehicles headlights to give a powerful beam of parallel light.
             (ii) Used as shaving mirrors to get large, magnified, virtual image.
             (iii) Used by dentist to see large image of patients teeth.


Q.39   If an object is placed at the focus of a concave mirror, where is the image formed ?
Sol.       At infinity.


Q.40   If an object is at infinity in front of concave mirror, where is the image formed ?
Sol.       At focus.


Q.41    For what position of an object a real and diminished image is formed by a concave mirror?
Sol.       When object is beyond C.


Q.42   A virtual, erect and magnified image of an object is to be produced with a concave mirror of focal length 12 cm. Which of the following object distance should be chosen for this purpose ?
(i) 10 cm  (ii) 15 cm   (iii) 20 cm
Give reasons for your choice.
Sol.     Object distance should be 10 cm. This is because a virtual erect magnified image can be formed by if the object is placed between P & F.


Q.43   A concave mirror has a focal length of 25 cm. At which of the following distance should a person hold his face from this concave mirror so that it may act as a shaving mirror ?
(a) 45 cm    (b) 20 cm    (c) 25 cm   (d) 30 cm
Sol.     A shaving mirror gives erect, virtual, magnified image only if object distance is less than the focal length. So answer is (b) 20 cm.


Q.44    An object is placed at the following distances from a concave mirror of focal length 15 cm, turn by turn :
(a) 35 cm      (b) 30 cm     (c) 20 cm    (d) 10 cm
Which position of the object will produce :
(i) a magnified real image ?
(ii) a magnified virtual image ?

(iii) a diminished real image ?
(iv) an image of same size as the object ?

Sol.

(i) 20 cm (object should be between F & C)
(ii) 10 cm (object should be between P & F)
(iii) 35 cm (Object should be beyond C or 2F)
(iv) 30 cm (Object should be at C)


Q.45    Name the type of mirror which is used in the headlights of a car. Why is it used for this purpose ?
Sol.        Concave mirror are used so as to give a powerful beam of parallel rays of light.


Q.46    Which type of mirror is used in solar furnace ? Support your answer with reason.
Sol.        Concave mirror. The parallel beam of sun’s rays will converge at the focus of the minor, thereby increasing the temperature.


Q.47    Name the type of mirror used by dentists. How does it help ?
Sol.        Concave mirror is used. It gives a magnified, erect image of the patient’s teeth.


Q.48    Explain why, concave mirrors are used as shaving mirrors.
Sol.      When a shaving mirror is kept at a distance of less than its focal length, an erect, magnified image of the face is seen. This makes shaving easier.


Q.49    An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.
                  (a) Draw a ray diagram for the formation of image.
                  (b) Calculate the image distance.
                  (c) State two characteristics of the image formed.
Sol.       (a) 32
                 (b) {1 \over f} = {1 \over v} + {1 \over u}
                        {1 \over { - 20}} = {1 \over v} + {1 \over { - 10}}
                        Therefore {1 \over v} = {1 \over { - 20}} + {1 \over {10}} = {{ - 1} \over {20}} + {2 \over {20}} = {1 \over {20}}
                        Therefore v = 20 cm
                 (c) Image is virtual and rest.


Q.50   What is the position of an image when an object is placed at a distance of 20 cm from a concave mirror of focal length 20 cm ?
Sol.         {1 \over f} = {1 \over v} + {1 \over u}
                   {1 \over { - 20}} = {1 \over v} + {1 \over { - 20}}
                   Therefore {1 \over v} = 0
                   Or v = \infty
                   Image is formed at infinity


Q.51   If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of the image.
Sol.        {1 \over f} = {1 \over v} + {1 \over u}
                 {1 \over { - 12}} = {1 \over v} + {1 \over { - 36}}
                 Therefore {1 \over v} = {{ - 1} \over {12}} + {1 \over {36}} = {{ - 1} \over {18}}
                 Therefore v = – 18 cm
                  {{h'} \over h} = {{ - v} \over u}
                  Therefore h' = - {{\left( { - 18} \right)} \over { - 36}} \times 10 = - 5\,cm
                  The image is real inverted at 18 cm, in front of mirror and of height 5 cm.


Q.52   Describe the nature of the image formed when the object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm.
Sol.       The focal length of the concave lens = 10 cm. The object is placed as 20 cm i.e. at C. So the image will be formed at 20 cm. It will be real inverted and of the same size as object -


Q.53   At what distance from a concave mirror of focal length 10 cm should an object be placed, so that its real image is formed 20 cm from the mirror ?
Sol.        f = – 10 cm, v = – 20 cm
                 {1 \over u} = {1 \over f} - {1 \over v}
                  = - {1 \over {10}} - \left( {{{ - 1} \over {20}}} \right) = - {1 \over {20}}
                 Therefore u = 20 cm


Q.54   If the magnification of a body of size 1 m is 2, what is the size of the image ?
Sol.         m = {{h'} \over h}
                  Therefore h’ = m × h = 1 × 2 = 2 m


Q.55   At what distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get and erect image 6 cm tall ?
Sol.         f = – 10 cm, h = 2 cm ; h’ = 6 cm
                  {{ - v} \over u} = {{h'} \over u} = {6 \over 2} = 3
                  Therefore v = – 3 u.
                  {1 \over f} = {1 \over v} + {1 \over u}
                  {1 \over { - 10}} = {1 \over { - 3u}} + {1 \over u} = {2 \over {3u}}
                  Therefore u = – 6.6 cm


Q.56   When an object is placed at a distance of 15 cm from a concave mirror, its image is formed at 10 cm in front of the mirror. Calculate the focal length of the mirror.
Sol.       u = – 15 cm ; v = – 10 cm
                {1 \over f} = {1 \over v} + {1 \over u}
                 = {{ - 1} \over {10}} - {1 \over {15}} = {{ - 1} \over 6}
                Therefore f = – 6 cm


Q.57   An object 3 cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 4.5 cm high :
               (i) What is the focal length of the mirror ?
              (ii) What is the position of image ?
             (iii) Draw a ray-diagram to show the formation of image.
Sol.       h = 3 cm ; u = – 8 cm , h’ = 4.5 (image is vertical so it is erect)
                {{ - v} \over u} = {{h'} \over n}
                v = - u \times {{h'} \over n}
                 = - \left( { - 8} \right) \times {{4.5} \over 3} = 12 cm
             (i) {1 \over f} = {1 \over v} + {1 \over u}

                        = {1 \over {12}} + {1 \over { - 8}} = {{ - 1} \over {24}}
                      Therefore f = – 24 cm
                 (ii) v = 12 cm
                 (iii) 33


Q.58   A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror :
              (i) Calculate the image distance.
             (ii) What is the focal length of the mirror ?
Sol.       h’ = – 4 cm, h = 1 cm ; u = – 20 cm
                (i) {v \over u} = {{ - h'} \over n}
                     {v \over { - 20}} = - {{\left( { - 4} \right)} \over 1} = – 80 cm
                (ii) {1 \over f} = {1 \over v} + {1 \over u}
                       = {1 \over { - 80}} + {1 \over { - 20}} = {{ - 1} \over {16}}
                      Therefore f = – 16 cm


Q.59   An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained ? Find the size and nature of image.
[Hint. Find the value of image distance (v) first. The screen Should be placed from the mirror at a distance equal to image distance]
Sol.         h = 7 cm , u = – 27 cm ; f = – 18 cm
                  {1 \over v} = {1 \over f} - {1 \over u}
                   = {{ - 1} \over {18}} - \left( {{{ - 1} \over {27}}} \right) = {{ - 1} \over {54}}
                  Therefore v = – 54 cm
                  The screen should be placed 54 cm in front of the mirror.
                   {{h'} \over h} = {{ - v} \over u}
                    = - {{\left( { - 54} \right)} \over { - 27}} = - 2
                   h' = 7 \times - 2 = - 14\,cm
                   Image is real, inverted size 14 cm.


Q.60   State the relation between object distance, image distance and focal length of a mirror.
Sol.      {1 \over f} = {1 \over v} + {1 \over u}


Q.61   What is the ratio of the height of an image to the height of the object known as ?
Sol.       Magnification


Q.62    Define linear magnification produced by a mirror.
Sol.        Linear magnification is defined as the ratio of height of image to height of object.


Q.63   Write down a formula for the magnification produced by a concave mirror in terms of image distance and object distance.
Sol.         m = {{ - v} \over u}


Q.64   Write the mirror formula. Give the meaning of each symbol which occurs in it.
Sol.        {1 \over f} = {1 \over v} + {1 \over u}
                  Where f is focal length
                   v is image distance
                   u is object distance


Q.65     Which mirror has a wider field of view?
Sol.         Convex mirror


Q.66    A man standing in front of a special mirror finds his image having a very small head, a fat body and legs of normal size. What are the shapes of three parts of the mirror?
Sol.         Convex mirror, concave mirror and plane mirror.


Q.67   If you want to see an enlarged image of your face, state whether you will use a concave mirror or a convex mirror.
Sol.         Concave mirror


Q.68     Which mirror always produces a virtual, erect and diminished image of an object?
Sol.         Convex mirror


Q.69     Which mirror is used as a rear- view mirror in vehicles? Why is it used for this purpose?
Sol.         Convex mirror. It gives erect image and wide field of view.


Q.70    Why does a driver prefer to use a convex mirror as a rear- view mirror in a car?
Sol.         As it gives a wide view and erect image.


Q.71    An object is placed at a long distance in front of a convex mirror of radius of curvature 30 cm. State the position of its image.
Sol.        Virtual erect , diminished formed behind the mirror.


Q.72     Name the spherical mirror which can produce a real and diminished image of an object.
Sol.         Concave mirror


Q.73    One wants to see an enlarged image of an object in a mirror. What type of mirror should one use?
Sol.         Concave mirror


Q.74    (a) Draw a ray diagram to show the formation of image in a convex mirror. What happens to the image when the object is moved away from the mirror?
                 (b) State three characteristics of the image formed by a convex mirror.
Sol.       (a) 


Q. 73

       The image size decreases and moves towards F.


Q.75      Name a mirror that can give an erect and enlarged image of an object.
Sol.          Concave mirror.


Q.76   An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature.
Sol.         u = –5 cm , f = 10 cm
               {1 \over v} = {1 \over f} - {1 \over u}
                = {1 \over {10}} - \left( { - {1 \over 5}} \right) = {3 \over {10}}
               Therefore v = 3.33 cm
               m = - {v \over u} = {{ - {3 \over {10}}} \over { - 5}} = 0.6
               Image is virtual


Q.77   An object is placed at a distance of 10 cm from a convex mirror of focal length 5 cm.
                (i) Draw a ray- diagram showing the formation of image.
                (ii) State two characteristics of the image formed.
                (iii) Calculate the distance of the image from mirror.
Sol.      (i)

Q. 76 Schand

                    (ii) Erect virtual
                   (iii) {1 \over v} = {1 \over f} - {1 \over u}
                          = {1 \over 5} - \left( {{{ - 1} \over {10}}} \right) = {3 \over {10}}
                          v = 3.3 cm


Q.78   An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.
Sol.          u = – 6 cm , f = 12 cm
                   {1 \over v} = {1 \over f} - {1 \over u}
                    = {1 \over {12}} - \left( {{{ - 1} \over 6}} \right) = {1 \over 4}
                    v = 4 cm
                   m = {{ - v} \over u} = {{ - 4} \over { - 6}} = {2 \over 3}
                   Image is at 4 cm behind the mirror, virtual, diminished.


Q.79  An object placed 20 cm in front of a mirror is found to have an image 15 cm (a) in front of it, (b) behind the mirror. Find the focal length of the mirror and the kind of mirror in each case.
Sol.     (a)  u = – 20 cm ; v = – 15 cm
                     {1 \over f} = {1 \over v} + {1 \over u}
                      = {{ - 1} \over {15}} - {1 \over {20}} = {{ - 7} \over {60}}
                     Therefore f = {{ - 60} \over 7}cm - Concave mirror.
           (b)  u = –20 cm ; v = + 15 cm

                      {1 \over f} = {1 \over {15}} - {1 \over {20}} = {1 \over {60}}
                      Therefore f = 60 cm ; Convex mirror


Q.80   An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.
Sol.          h = 2.5 cm u = – 25 cm f = 20 cm
                   {1 \over v} = {1 \over f} - {1 \over u}
                    = {1 \over {20}} - \left( {{1 \over {25}}} \right) = {9 \over {100}}
                   Therefore v = 11.1 cm
                    h' = - h \times {v \over u} = - 2.5 \times {{11.1} \over { - 25}} = 1.11\,cm
                    Image is erect, at 11.1 cm behind the mirror, and height 1.11 cm.


Q.81   A convex mirror used as a rear- view mirror in a car has a radius of curvature of 3m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image?
Sol.        f = {R \over 2} = {3 \over 2}m\,;\,u = - 5\,m
                  {1 \over v} = {1 \over f} - {1 \over u}
                   = {2 \over 3} - \left( {{{ - 1} \over 5}} \right) = {{13} \over {15}}
                  Therefore v = 1.15 cm
                  Image is virtual and erect.


Q.82   What is refraction of light? Draw a diagram to show the refraction of light.
Sol.       The bending of light when it passes from one medium to another is called refraction.

36


Q.83   If a ray of light goes from a rarer medium to a denser medium, will it bend towards the normal or away from it?
Sol.       Towards the normal


Q.84   If a ray of light goes from a denser medium to a rarer medium, will it bend towards the normal or away from the normal?
Sol.       Away from normal


Q.85   A beam of light travelling in a rectangular glass slab emerges into air. Draw a ray- diagram indicating the change in its path.
Sol. 8


Q.86   A beam of light travelling in air is incident on water. Draw a ray- diagram indicating the change in its path in water.
Sol. 37


Q.87   A ray of light travelling in water emerges into air. Draw a ray- diagram indicating the change in its path.
Sol.38


Q.88    A ray of light travelling in air is incident on a parallel- sided glass slab (or rectangular glass slab). Draw a ray- diagram indicating the change in its path in glass.
Sol.       

8


Q.89   A ray of light travelling in glass emerges into air. State whether it will bend towards the normal or away from the normal.
Sol.       Away from normal


Q.90  A ray of light goes from air into glass. Will it bend away from the normal or towards the normal?
Sol.     Toward normal


Q.91  A ray of light travelling in air is incident on a rectangular glass block and emerges out into the air from the opposite face. Draw a ray- diagram to show the complete path of this ray of light.
Sol.    


8


Q.92   A ray of light travelling in air enters obliquely into water. Does the ray of light bend towards the normal or away from the normal? Why?
Sol.      It bends towards normal, as water is optically denser than air.


Q.93   A ray of light goes from water into air. Will it bend towards the normal or away from the normal?
Sol.       Away from normal.


Q.94   Define the term ‘refractive index’.
Sol.       The refractive index of a transparent medium is the ratio of the speed of light in vacuum to that in the medium.
             
n = {c \over v}


Q.95   What is the unit of refractive index?
Sol.       Refractive index has no units.


Q.96   Which has higher refractive index, water or glass?
Sol.       Glass.


Q.97  Refractive indices of carbon disulphide and ethyl alcohol are 1.63 and 1.36 respectively. Which is optically denser?
Sol.      Carbon disulphide.


Q.98   State the law of refraction of light.
Sol.       Law of refraction of light.
                (i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the 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}} = Cons\tan t = refractive index of second medium w.r.t. the first medium.
                       Where i is angle of incidence
                        r is angle of refraction.


Q.99   Define Snell’s law of refraction.
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 colour and for the given pair of media. This law is known as snell’s law.
{{\sin \,i} \over {\sin \,r}} = Cons\tan t = refractive index of second medium w.r.t. the first medium.
Where i is angle of incidence
r is angle of refraction.


Q.100  Write the relation between the angle of incidence and angle of refraction for a medium.
Sol.        {{\sin \,i} \over {\sin \,r}} = refractive index of second medium w.r.t. first medium.


Q.101  What name is given to the ratio of since of angle of incidence to the sine of angle of refraction?
Sol.        Refractive index


Q.102  The speed of light in water is 2.25 \times {10^8} m/s. If the speed of light in vacuum be 3 \times {10^8}\,m/s, calculate the refractive index of water.
Sol.    {\eta _{wa}} = {{{v_{air}}} \over {{v_{water}}}} = {{3 \times {{10}^8}m/s} \over {2.25 \times {{10}^8}m/s}} = 1.33


Q.103   A ray of light is incident on a glass slab at an angle of incidence of 60°. If the angle of refraction be 32.7°, calculate the refractive index of glass. (Given : sin 60°= 0.866, and 32.7°=0.540)
Sol.        {n_{ga}} = {{\sin \,i} \over {\sin \,r}} = {{\sin \,60} \over {\sin \,32.7}} = 1.60


Q.104   The refractive index of diamond is 2.42. What is the meaning of this statement?
Sol.         It means that the speed of light in air is 2.42 times greater that the speed of light in diamond.


Q.105   If the refractive index of diamond for light going from air to diamond be 2.42, what will be the refractive index for light going from diamond to air?
Sol.          r i (of air to diamond = 1) r.i. (of diamond to air).
                   or {n_{21}} = {1 \over {{n_{12}}}}
                   Therefore {n_{12}} = {1 \over {{n_{21}}}} = {1 \over {2.42}} = 0.41


Q.106   What is a lens? Distinguish between a convex lens and a concave lens.
Sol.        A lens is a piece of transparent material bounded by two spherical surfaces.
                 A concave lens is thick at the centre and thin at the edges.
                 A concave lens is thin at the centre and thick at the edges.


Q.107   With the help of a diagram, explain the terms focal length and principal axis of a convex lens.
Sol.

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The principal axis is an imaginary line passing through the optical centre and the two centre of curvature of the  lens.  
 Rays parallel to the principal axis converge at a point on the principal axis. Which is called focus. The distance  between  optical centre and focus is called focal length.


Q.108   Explain with the help of a diagram, why the convex lens is also called a converging lens.
Sol.9

Since all parallel rays meet at a point it is called as converging lens.


Q.109  Explain with the help of a diagram, why the concave lens is also called a diverging lens.
Sol. 10

All the rays parallel to the principal axis diverge, it is called a diverging lens.


Q.110  Define the principal focus of a convex lens.
Sol.      When rays parallel to the principal axis fall on a convex lens, they are refracted and converge at a point on the principal axis. The point is known as the principal focus.


Q.111  Draw a labelled diagram to show the : (i) optical centre (ii) Principal axis, (iii) Principal focus, and (iv) Focal length of a convex lens.
Sol. 11

(i) O is the optical centre.
(ii) Line passing through 2F1 to 2F2 is the principal axis.
(iii) F2 is the principal focus.
(iv) Distance between O to F2 is the focal length .


Q.112  Give the usual name for the following :
A point inside a lens through which the light passes undeviated.
Sol.       Optical centre


Q.113  Which of the two is a diverging lens : Convex or concave?
Sol.        Concave lens


Q.114  State whether a concave lens has a virtual focus or real focus.
Sol.       Virtual focus


Q.115  Define the principal focus of a concave lens with the help of a diagram.
Sol. 12

Rays parallel to the principal axis diverge after passing through a concave lens. The point from which they appear to diverge is called the principal axis of a concave lens.


Q.116  Name the lens which can concentrate sun’s light rays to a point and burn a hole in a piece of paper.
Sol.        Convex lens.


Q.117  Draw a ray diagram to show the formation of a real magnified image by a convex lens. (In your sketch the position of object and image with respect to the principal focus of lens should be shown clearly).
Sol. 13


Q.118  A 1 cm high object is placed at a distance of 2f from a convex lens. What is the height of the image formed?
Sol.        1 cm


Q.119   If the image formed by a convex lens is of the same size as that of the object, what is the position of the image with respect to the lens?
Sol.         2 f


Q.120  State with the help of a ray- diagram the formation of image of a finite object placed in front of a convex lens between f and 2f. Give two characteristics of the image so formed.
Sol.14

Image is real and magnified.


Q.121   An object is placed at the following distances from a convex lens of focal length 15 cm.
(a) 35 cm                       (b) 30 cm                           (c) 20 cm                           (d) 10 cm
Which position of the object will produce :
(i) a magnified real image ?
(ii) a magnified virtual image?
(iii) a diminished real image?
(iv) an image of same size as the object?
Sol.      (i) 20 cm     (ii) 10 cm   (iii) 35 cm   (iv) 30 cm


Q.122  If an object is placed at the focus of a convex lens, where is the image formed?
Sol.       At infinity


Q.123  For what position of an object, a virtual image is formed by a convex lens?
Sol.       When the object is placed between focus to optical certic.


Q.124 Where should an object be placed in front of a convex lens, so as to obtain its magnified erect image?
Sol.        When the object is placed between focus to optical certic.


Q.125  Where should an object be placed in front of a convex lens, so as to obtain its real, magnified and inverted image?
Sol.        When the object is between F + 2 F.


Q.126  For what position of an object, a real, diminished image is formed by a convex lens?
Sol.       When the object is beyond 2 F.


Q.127 If an object is at a considerable distance (or infinity) in front of a convex lens, where is the image formed?
Sol.      At focus


Q.128  Describe with the help of a ray diagram the nature, size and position of the image formed when an object is placed in front of a convex lens between focus and optical centre. State three characteristics of the image formed.
Sol. 15

Image is virtual erect and magnified.


Q.129   An object is placed at a distance equal to 2f in front of a convex lens. Draw a labelled ray diagram to show the formation of image. State two characteristics of the image formed.
Sol.16

Image is real and inverted.


Q.130  Describe with the help of a ray-diagram, the size, nature and position of the image formed by a convex lens when an object is placed beyond 2f in front of the lens.
Sol.       

 

16

   Image is real and inverted.


Q.131  Where should an object be placed in order to use a convex lens as a magnifying glass?
Sol.       When the object is between O + F


Q.132  A convex lens has a focal length of 10 cm. At which of the following position should an object be placed so that this convex lens may act as a magnifying glass?
(a) 15 cm    (b) 7 cm   (c) 20 cm   (d) 25 cm
Sol.        7 cm


Q.133  Which one of the following material cannot be used to make a lens?
(a) Water  (b) Glass (c) Plastic (d) Clay
Sol.        Clay


Q.134  Write the formula for a lens connecting image distance (v), object distance (u) and the focal length (f). Explain the sign convention you use.
Sol.       {1 \over f} = {1 \over v} - {1 \over u}
                Sign convention :
                1. All measurement are taken from the optical centre.
                2. The object is placed on the left of lens.
                3. All distance measured in the direction of light are taken positives, and in the opposite direction as negative.
           4. Distances measured perpendicular and above the principal axis are taken as positive and below are taken negative.


Q.135 An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.
Sol.       h = 4 cm
                u = - 10 cm
                f = 20 cm
                {1 \over v} = {1 \over f} + {1 \over u}
                 = {1 \over {20}} + {1 \over { - 10}} = {{ - 1} \over {20}}
                u = – 20 cm
                {{h'} \over h} = {v \over u}
                Therefore h' = 4 \times {{ - 20} \over { - 10}} = 8\,cm
                The image is at 20 cm in font of lens, virtual erect and 8 cm height.


Q.136 A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.
Sol.       f = 5 cm
                v = –25 cm (as image is virtual)
                {1 \over u} = {1 \over v} - {1 \over f}
                 = {{ - 1} \over {25}} - {1 \over 5} = {{ - 6} \over {25}}
                Therefore u = {{ - 25} \over 6}
                 m = {v \over u} = {{ - 25} \over {{{ - 25} \over 6}}} = 6
             Ans : 6.


Q.137  Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.
Sol.       h = 5 cm, u = - 10 cm, f = 6 cm
                {1 \over v} = {1 \over f} + {1 \over u}
                 = {1 \over 6} + {1 \over { - 10}} = {1 \over {15}}
                 v = 15 cm
                The image is real
                {{h'} \over h} = {v \over u}
                h' = 5 \times {{15} \over { - 10}} = - 7.5
                Image is inverted, at 15 cm behind the lens.


Q.138  Calculate the focal length of a convex lens which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.
Sol.        v = –50 cm , u = – 20 cm
                 {1 \over f} = {1 \over v} - {1 \over u}
                  = {{ - 1} \over {50}} - \left( {{{ - 1} \over {20}}} \right)
                 f = 33.3 cm


Q.139  An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm.
           
(i) What is the nature of image?
                  (ii) What is the position of image?
Sol.       u = – 100 cm , f = 40 cm
                 (i) Object is beyond 2f so the image is real and inverted.
                 (ii) {1 \over v} = {1 \over f} + {1 \over u}
                        = {1 \over {40}} - {1 \over {100}}
                        v = 66.6 cm


Q.140  A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens.
Sol.      m = – 3 , u = – 15 cm
               m = {{ - v} \over u}
               Therefore v = + (–3) × (–15) = 45 cm
               {1 \over f} = {1 \over v} - {1 \over u}
                = + {1 \over {45}} - \left( {{{ - 1} \over {15}}} \right) = {4 \over {45}}
               f = 11.2 cm


Q.141 A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?
Sol.      f = 5 cm ; v = 20 cm
               {1 \over u} = {1 \over v} - {1 \over f}
                = {1 \over {20}} - {1 \over 5} = {{ - 3} \over {20}}
               u = – 6.6 cm


Q.142  What is the difference between the mirror formula and the lens formula?
Sol.      Mirror formula {1 \over v} + {1 \over u} = {1 \over f}
               Lens formula {1 \over v} - {1 \over u} = {1 \over f}


Q.143  Write down the magnification formula for a lens in terms of object distance and image distance. How does it differ from the corresponding formula for a mirror?
Sol.       Magnifications of lens m = {v \over u}
                Magnifications of mirror m = {{ - v} \over u}


Q.144   An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.
Sol.      h = 5 cm, u = –25 cm , f = 10 cm
               {1 \over v} = {1 \over f} + {1 \over u}
                = {1 \over {10}} - {1 \over {25}} = {3 \over {50}}
               v = 16.6 cm
               {{h'} \over h} = {v \over u}
               Therefore h' = {{5 \times {{50} \over 3}} \over { - 25}} = - 3.3cm
           
Position 16.6 cm, size 3.3 cm, real inverted.


Q. 143 Schand


Q.145  A spherical mirror and a spherical lens each have a focal length of, –15 cm. The mirror and the lens are likely to be :
(i) Both concave
(ii) Both convex

(iii)The mirror is concave but the lens is convex.
(
iv)The mirror is convex but the lens is convex.

Sol.       (i) Both concave.


Q.146  If the image formed by a lens is always diminished and erect, what is the nature of the lens?
Sol.      Concave lens


Q.147 An object lies at a distance of 2f from a concave lens of focal length f. Draw a ray–diagram to illustrate the image formation.
Sol. 

41


Q.148  Show by drawing a ray- diagram that the image of an object formed by a concave lens is virtual, erect and diminished.
Sol. 17


Q.149  What kind of lens can form :
            (a) An inverted magnified image?                                                      
          (b) An erect magnified image?

          (c) An inverted diminished image?                                                
          (d) An erect diminished image ?

Sol.      (a) Convese lens
               (b) Convex lens
               (c) Convese lens
               (d) Concave lens


Q.150 Give the position, size and nature of image formed by a concave lens when the object is placed :
                 (a) anywhere between optical centre and infinity.              
              
(b) At infinity

Sol.      (a) Between O and F                                                                
            (b) At F


Q.151 An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.
Sol.       u = 4 cm ; f = –12 cm
                {1 \over v} = {1 \over f} + {1 \over 4}
                 = {{ - 1} \over {12}} + \left( {{{ - 1} \over 4}} \right) = {{ - 1} \over 3}
                V = – 3 cm
                Image is 3 cm is front of lens, virtual erect.


Q.152  A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray- diagram.
Sol.      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

42


Q.153  An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.
Sol.      u = – 60 cm v = – 20 cm
               {1 \over f} = {1 \over v} - {1 \over u}
                = {{ - 1} \over {20}} - \left( {{{ - 1} \over {60}}} \right) = {{ - 2} \over {60}}
               f = – 30 cm
               The lens in a diverging lens as f is negative.


Q.154 A concave lens of 20 cm focal length forms an image 15 cm from the lens. Compute the object distance.
Sol.        f = – 20 cm , v = – 15 cm
                 {1 \over u} = {1 \over v} - {1 \over f}
                  = {{ - 1} \over {15}} - \left( {{{ - 1} \over {20}}} \right) = {{ - 1} \over {60}}
                 Therefore u = – 60 cm


Q.155  An object is 2 m from a lens which forms an erect image one–fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?
Sol.        u = – 2m, h' = {h \over 4}
                 {v \over u} = {{h'} \over h}
                {1 \over f} = {1 \over v} - {1 \over u} = {{ - 1} \over 1} - \left( {{{ - 1} \over 2}} \right) = {{ - 3} \over 2}
                 Therefore f = – 0.66 m
                  Type of lens – Concave lens.


Q.156  A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it form an image at 10 cm from the lens? Also find the magnification produced by the lens.
Sol.       f = –15 m ; v = –10 cm
                 = {{ - 1} \over {10}} - \left( {{{ - 1} \over {15}}} \right) = {{ - 1} \over {30}}
                Therefore u = – 30 cm
                m = {v \over u} = {{ - 10} \over { - 30}} = 0.33


Q.157  Define 1 dioptre power of a lens.
Sol.       One dioptre is the power of lens whose focal length is 1m.


Q.158  A diverging lens has a focal length of 3 cm. Calculate the power and give the units of power.
Sol.       P = {1 \over f} = {{ - 1} \over {3400}} = - 33.3\,D


Q.159  The focal length of a convex lens is 25 cm. What is its power?
Sol.        P = {{100} \over {25}} = 4D


Q.160  The power of a lens is , – 2 D. What is its focal length?
Sol.        f = {1 \over P} = {{ - 1} \over 2} = - 0.50\,m


Q.161  What do you understand by the power of a lens? Define also its unit.
Sol.      Power of a lens is the degree of its convergence or divergence. It is the recipeocal of its focal length in metres. Its unit is dioptres.


Q.162  What is the SI unit of power of a lens?
Sol.       SI unit of power is dioptre. One dioptre is the power of lens whose focal length is 1m.


Q.163  An object of height 4 cm is placed at a distance of 15 cm in front of a concave lens of power, – 10 dioptres. Find the size of the image.
Sol.       f = {1 \over P} = {{ - 100} \over {10}} = - 10\,cm
                 H = 4 cm ; u = –15 cm
                 {1 \over v} = {1 \over f} + {1 \over u}
                  = {{ - 1} \over {10}} - {1 \over {15}} = {{ - 1} \over 6}
                 Therefore v = – 6 cm
                 {{h'} \over h} = {v \over u} = {{ - 6} \over { - 15}}
                 Therefore h' = {6 \over {15}} \times 4 = 1.6\,cm


Q.164  Which type of lens has a positive power?
Sol.        Convex lens


Q.165  How is the power of a lens related to its focal length?
Sol.        P = {1 \over f} when f is in metres.


Q.166  Which of the two has a greater power : a lens of short focal length or a lens of large focal length?
Sol.        Short focal length


Q.167   Name the physical quantity whose unit is diopter.
Sol.         Power


Q.168   A lens has a focal length of –10 cm. What is the power of the lens and what is its nature ?
Sol.         P = {{ - 100} \over {10}} = - 10D ; concave lens


Q.169  The power of a lens is, +4 D. What kind of lens is it and what is its focal length?
Sol.        P = + 4 D, so it is a convex lens
                 f = {1 \over 4} = 0.25{\mkern 1mu} m


Q.170  What is the nature of a lens having a power of – 5 D?
Sol.        Diverging lens.


Q.171  Which type of lens has a negative power?
Sol.       Concave lens.


Q.172  The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm, calculate the focal length of lens Y .
Sol.       P = {P_1} + {P_2}
                5 = {{100} \over {15}} + {P_2}
                Therefore {P_2} = 5 - {{20} \over 3} = {{ - 5} \over 3}
             Therefore {f_z} = {{ - 100} \over {{5 \over 3}}} = - 60cm


Q.173  A convex lens of power 5 D and a concave lens of power 7.5 D are placed in contact with each other. What is the power of this combination of lenses?
Sol.      P = {P_1} + {P_2} = + 5 – 7.5 = –2.5 D


Q.174  A convex lens of focal length 10 cm and a concave lens of focal length 20 cm are placed in close contact with each other.
(a) What is the power of this combination ?
(b) What is the focal length of this combination?

Sol.       {1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}}
                 = {1 \over {10}} - {1 \over {20}} = {1 \over {20}}
                (a) P = {{100} \over f} = {{100} \over {20}} = 5\,D
                (b) f = 20 cm


Q.175  What is the power of a convex lens of focal length 0.5 m?
Sol.       P = {1 \over f} = {1 \over {0.5}} = 2D


Q.176  A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Sol.        f = {1 \over p} = {1 \over {1.5}} = 0.66m
              
It is a converging lens.


Q.177  A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another.
(a) What is the power of this combination ?
(b) What is the focal length of this combination?
Sol.       {1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}}
                 = {1 \over {25}} - {1 \over {10}} = {{ - 3} \over {50}}
                P = {1 \over f} = {{ - 3} \over {50}} \times 100 = - \,6D
                f = {{ - 50} \over 3} = - 16.6\,cm

 



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