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Question

6. a.An object 4 cm in size is placed at a distance of 25 cm from a concave mirror of focal
length 15 cm. Find the position, nature and height of the image.
b 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. Calculate the image distance. What is the focal
length of the mirror?

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Solution

Given that f = -15 cm; u = -25 cm ; ho= 4 cm; v = ?
The mirror formula is 1/f = 1/u + 1/v = 1/-15 = 1/-25 + 1/v
1/v = 1/25 - 1/15
1/v = (3-5)/ 75 =-2/ 75
v =-37.5 cm
The screen should be placed 37.5 cm from the pole of mirror and the image is real.
Magnification, m = -v/u =hi/ho= hi / 4 = 37.5 / 25
hi = (37.5 / 25 )x 4 = -6 cm
So the image is enlarged and inverted.


b)# Converging mirror is actually concave mirror!

hi = -4cm(real and inverted.)
ho = 1cm
u=-20cm
(a) v=?

hi/ho=-v/u

where,
hi=height of the image
ho=height of the object
v=distance of the image from the mirror
u=distance of the object from the mirror

-4/1=-v/-20
4 = v/20
v = -80cm

Let focal length of the mirror to be 'f'

1/f = 1/v + 1/u

1/f = 1/-80 + 1/(-20)
1/f = 1/-80 - 1/ 20
1/f = -5/80
f = -80/5cm = -16 cm

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