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A concave lens of focal length $15 cm$ forms an image $10 cm$ from the lens. The image formed

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inverted

erect

virtual

none of these

answer is B.

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Detailed Solution

The image formed is erect.
Given,
Image distance,

v = - 10 cm

and focal length,

f = - 15 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}{(- 30)}

\Rightarrow u = - 30 cm

Hence, the distance of the object should be

- 30 cm

.
It is known that the magnification

(m)

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 magnification

(m)

\Rightarrow m = \frac{v}{u}

\Rightarrow m = \frac{- 10}{- 30}

\Rightarrow m = 0.33

Hence, the magnification for the image formed is

0.33 ModerateFormula basedApplication

.
The positive sign indicates that the image formed is erect.
$Question Image$

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