The magnification of an image formed by a lens is -1. If the image's distance from the lens's optical centre is 25 cm on the opposite side of the object. Where is the object placed? Find the nature and focal length of the lens. Where would the image be formed if the object is displaced 15 cm towards the optical centre of the lens?

The nature of the lens is convex with a focal length of

+ 12.5 cm

, and the image is formed at

- 50 cm

if the object is displaced

15 cm

towards the optical centre of the lens.
Given,
Magnification,

m = - 1

Image distance,

v = 25 cm

For real image,

m = - 1

;
Here,

m = \frac{v}{u}

\Rightarrow

\frac{v}{u} = - 1

\Rightarrow

u = - v

\Rightarrow

u = - 25 cm

By using lens formula, we get

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

\Rightarrow \frac{1}{f} = \frac{1}{25} - \frac{1}{- 25}

\Rightarrow \frac{1}{f} = \frac{1}{25} + \frac{1}{25}

\Rightarrow \frac{1}{f} = \frac{2}{25}

\Rightarrow f = + 12.5 cm

Thus, the positive focal length shows that the given lens is convex with a focal length of

12.5 cm

. If the object is shifted

15 cm

towards the optical centre of the lens, the object is now placed at a distance of

25 - 15 = 10 cm

from the optical centre.
Therefore

u = - 10 cm

, and

f = + 12.5 cm

.
Using lens formula again,

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

\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u}

\Rightarrow \frac{1}{v} = \frac{2}{25} + \frac{1}{- 10}

\Rightarrow \frac{1}{v} = \frac{4 - 5}{50}

\Rightarrow \frac{1}{v} = \frac{- 1}{50}

\Rightarrow

v = - 50 cm

A negative sign indicates that the image will be formed on the same side as that of the object.

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