Question

SOLVE THE FOLLOWING NUMERICALS.

1. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it where is
the image located?

2. 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.
a) Calculate the image distance.
b) What is the focal length of the mirror?

3. A doctor has prescribed a corrective lens of power-4.5 D to a person .find the focal length of the lens .the
person is suffering from which defect?

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Answer 1

Answer:

Explanation:

1) Here, linear magnification (m) = - 3 (Negative sign for real image, which is inverted) Object distance. Image distance = The image is located at 30 cm in front of the mirror.

2)  Converging mirror is a concave mirror

hi = 4cm

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 = -3/80

f = -80/3 cm

3) Negative sign of power of lens indicates that focal length of the lens used for correction is negative. This means that concave lens is used for correction. Concave lens is used for the correction of myopia and hence, the person is suffering from myopia.

Focal length = 1/P

f = 1/-4.5

f = 22.2 cm

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