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REFLECTION OF LIGHT Numericals CLASS 9 PHYSICS CHAPTER 7

ICSE CLASS9 PHYSICS CHAPTER 7 REFLECTION OF LIGHT Numericals

REFLECTION OF LIGHT

Exercise 7A
Numerical
1.A ray of light is incident on a plane mirror.Its reflected ray is perpendicular to the incident ray.Find
   the angle of incidence?
Answer:
   Angle of incidence (i) + Angle of reflection(r) = 90o  
   But, according to the laws of reflection, i = r 
   Therefore, 2 i = 90o
    i = r = 45o 

2. A man standing in front of a plane mirror finds his image at a distance 6 meter from himself. What
    is the distance of man from the mirror?
Answer:
    Distance between man and his image = 6 m
    Distance between man and mirror + distance between mirror and image = 6m
    But, Distance between man and mirror = distance between mirror and image 
    Therefore, distance of man from mirror = 6/2 = 3 m

3. An insect is sitting in front of a plane mirror at a distance 1m from it.
    (a) Where is the image of the insect formed?
    (b) What is the distance between the insect and its image?
Answer:
    (a) Image of the insect is formed 1m behind the mirror.
    (b) Distance between the insect and his image = 1 + 1 = 2 m
4. An object is kept at 60 cm in front of a plane mirror.If the mirror is now moved 25 cm away from 
    the object,how does the image shift from its previous position?
Answer:
    Distance of the object from the mirror = 60 cm.
    Therefore, image is formed at a distance 60 cm from the mirror, behind it.
    Thus, initial distance between the object and image = 60 + 60 = 120 cm
    If the mirror is moved 25 cm away from the object,the new distance of the
    object from the mirror = 60 + 25 = 85 cm
    The new image is now at a distance 85 cm from the mirror behind it.
    Thus, new distance of the image from the object = 85 + 85 = 170 cm
    Taking the position of the object as reference point, the distance between the
    two positions of the image = new distance of image from the object - initial
    distance of the image from the object = (170 - 120) cm = 50 cm
    Thus, the image shifts 50 cm away.
5. An optician while testing the eyes of a patient keeps a chart of letters 3 m behind the patient and
    asks him to see the letters on the image of chart formed in a plane mirror kept at distance 2 m
    in front of him.At what distance is the chart seen by the patient?
Answer:
    Distance between man and chart = 3 m
    Distance between man and mirror = 2 m
    Therefore, distance between chart and mirror = 5 m
    Now, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore,  
    the chart as seen by patient is (5 m + 2 m =) 7 m away.
Exercise 7 (B)
Numerical
1. State the number of images of an object placed between the two plane mirrors,formed in each case
    when the mirrors are inclined to each other at    
    (a) 90o and (b) 60o
Answer:
 
    (a) Angle between the mirrors, แ€ = 90o
           Now, n = 360o / แ€ = 360o / 90o = 4, which is even.
           Hence number of images formed will be (n-1); i.e., 4-1 = 3 images
    (b) Angle between the mirrors, แ€ = 60o

           Now, n = 360o / แ€ = 360o / 60o = 6, which is even.

 2. An Object is placed (i) asymmetrically (ii) symmetrically, between two plane mirrors inclined

     at an angle of 50o.Find the number of images formed? 

Answer:

     Angle between the mirrors, = 50o
     
     Now, n = 360o / แ€ = 360o / 50o = 7.2 7, which is odd.

     (i) When placed asymmetrically, number of images formed will be n, i.e. 7.

     (ii) When placed symmetrically, number of images formed will be (n-1);
      
           i.e. 7-1 = 6 images
Exercise 7(C)
Numerical
1. The radius of curvature of a convex mirror is 40 cm.Find its focal length?
Answer:
     Focal length = ½ (Radius of curvature) Or, f = 40/2 = 20 cm

2. The focal length of a concave mirror is 10 cm. Find its radius of curvature.
Answer:
    Radius of curvature = 2 focal length
    Or, R = 2f = 2 10 = 20 cm

3. An object of height 2 cm is placed at a distance 20 cm in front of a concave mirror of focal length
   12 cm. Find the position,size and nature of the image.
Answer:
   The image is 30 cm in front of the mirror, 3 cm high, real, inverted and magnified.


4. An object is placed at 4 cm distance in front of a concave mirror of radius of curvature 24 cm.Find 
    the position of image.Is the image magnified?
Answer:
    The image is 6 cm behind the mirror.
    Yes. the image is magnified.

5. At what distance from a concave of focal length 25 cm should an object be
    placed so that the size of image is equal to the size of the object?
Answer:
   The size of the image is equal to the size of the object if the object is placed at the centre of 
   curvature of a concave mirror.Hence, the object should be placed at 50 cm.

 6.An object 5 cm high is placed at a distance 60 cm in front of a concave mirror of focal length 10
    cm.Find (i) the position and (ii) size of the image.
    The position of the object is 12 cm in front of the mirror.
    Its size is 1 cm. 

7. A point light source is kept in front of a convex mirror at a distance of 40 cm.The focal length
    of the mirror is 40 cm. Find the position of the image.
Answer:
   The image is behind the mirror at a distance 20 cm.

8. When an object of height 1 cm is kept at a distance 4 cm from a concave mirror,its erect image
    of height 1.5 cm is formed at a distance 6 cm behind the mirror.
    Find the focal length of mirror 
Answer:
   A ray passing parallel to the principal axis passes through the focal point after reflection. Hence, the
   focal length is 12 cm.
9. An object of length 4 cm is placed in front of a concave mirror at a distance 30 cm.The focal
    length of mirror is 15 cm. (a) Where will the image form?
    (b) What will be the length of the image?
Answer:
    O=4 cm ,u=-30 cm ,f=-15 cm
    from mirror formula 1 / f= 1 / u + 1 / v 
                               1 / v=1 / f - 1 / u = 1 / -15 - 1 / -30 = 1 / 30 - 1 / 15 
                               1 / v = - 1 / 30 ,v = -30 cm
                               Hence the image is formed at a distance of 30 cm in front of the mirror.
    m= - v / u= I / O
    I = - Ov / u = -4* -30 / -30 = -4 cm
    Negative sign indicates inverted image.So I = 4 cm. The length of the image is 4 cm.

10. A concave mirror forms a real image of an object placed in front of it at a distance 30
      cm,of size three times the size of object. Find (a) the focal length of mirror 
      (b) position of image
Answer:
      u= -30 cm
      m= I / O = 30 / O = 3
      But , for real object m is negative
      m= -3
      m= - ( v / u ) , ∴ - (v / u ) = -3
      v= 3u
      v= 3 x -30 = -90 cm
      So the position of image is 90 cm in front of the mirror.
  
      From mirror formula,
      1/f = 1/v + 1/u
      ∴ 1/f = 1/ -90 + 1 / -30 = -(1 / 90 )- (1 / 30 )=( - 1 -3 ) / 90 = -4 / 90
      f = -22.5 cm    
11. A concave mirror forms a virtual image of size twice that of the object placed at a distance 5 cm
      from it.Find :(a) the focal length of the mirror (b) position of image.
Answer:
      m = I / O = 20 / 2 = 2
      m= -(v / u)
      ∴ -(v / u) = 2
      -v = 2 u
      u= -5 cm
      v = 10 cm
      So the position of image is 10 cm in behind the mirror.
      From mirror formula,
      1/f = 1/v + 1/u
      ∴ 1/f = 1/ 10 + 1 / -5 =(1 / 10) - (1 / 5)
      f= -10 cm       
12.The image formed by a concave mirror is of size double the size of object. How are u and v 
      related?
Answer:
      magnification is 
      m = I / O = (1/3 O) / O = 1 / 3
      A convex mirror always forms a virtual and upright image. So ,  
      m= -v / u
      -v / u =1 / 3
      v = - 1/ 3 u
      But , u is always negative
      v = 1 / 3 u
      u = -3v
13.The erect image formed by a concave mirror is of size double the size of object.How are u and v
      related?
Answer:
      magnification is 
      m = I / O = 2 O / O = 2
      m = - (v / u)
      -(v / u) = 2
      v = -2u
      But , u is always negative
      v = 2u
14.The magnification for a mirror is -3.How are u and v related?
Answer:
       Magnification of a mirror is
      m = - v / u 
      m = -3 
      -3 = - (v / u )
      v = 3 u
      But , u will always be negative
      v = -3 u

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