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A diverging lens has a focal length of $20 cm$ , with the height of the image formed as $4 cm$ from the optical centre of the lens. The image is formed $10 cm$ away from it. What is the size and distance of the object to be placed?

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8 cm, 20 cm

8 cm, - 20 cm

1 cm, 20 cm

1 cm, - 20 cm

answer is B.

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Detailed Solution

The size and distance of the object to be placed is

8 cm

and

- 20 cm

respectively.
Given
Image distance,

v = - 10 cm

,
Focal length,

f = - 20 cm

and [Negative: as the lens is concave or diverging]
Image height,

h' = 4 cm

.
Form the lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

Where

f

v

, and

u

are the focal length, image distance and object distance, respectively.
On arranging the terms, we get

\frac{1}{u} = \frac{1}{v} - \frac{1}{f}

On substituting the given values, we get the object distance as,

\Rightarrow \frac{1}{u} = \frac{1}{(- 10)} - \frac{1}{- 20}

\Rightarrow \frac{1}{u} = \frac{- 2 - (- 1)}{20}

\Rightarrow u = - 20 cm

Hence, the distance of the object is

- 20 cm

.
It is known that the magnification produced is the ratio of the image distance

(v)

to the object distance

(u)

. It can also be defined as the ratio of the image height

(h')

to the object height

(h)

. Mathematically,

m = \frac{v}{u} = \frac{h'}{h}

On substituting the given values, we get the object height as

(h)

\frac{v}{u} = \frac{h'}{h}

\Rightarrow h = \frac{h' u}{v}

\Rightarrow h = \frac{4 \times (- 20)}{- 10}

\Rightarrow h = + 8 cm

A positive sign indicates that the image is erect.

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