Question

A concave lens has focal length of 15 cm. At what distance should
the object from the lens be placed so that it forms an image at 10 cm
from the lens? Also, find the magnification produced by the lens.

Answer 1

Answer:

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. The image is real, inverted and enlarged in size. A concave lens of focal length 15 cm forms an image 10 cm from the lens.

Explanation:

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Answer 2

Answer:

Question:-

A concave lens has focal length of 15 cm. At what distance should

the object from the lens be placed so that it forms an image at 10 cm

from the lens? Also, find the magnification produced by the lens.

solution:-

A concave lens always forms a virtual, erect image on the same side of the object.

Given,

Image distance, v = -10 cm

Focal length, f = - 15 cm [f is -ve for a concave lens]

Object distance, u = ?

Now, using lens formula,

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

we have ,

$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} \\ \\ = \frac{1}{ - 10} - \frac{1}{ - 15} \\ \\ = \frac{ - 3 +2 }{30} \\ = - \frac{1}{30}$

i.e., u = - 30 cm

Thus the object should be placed at a distance of 30 cm from the lens on the left side.

Now,

Magnification,

$m = \frac{v}{u} = \frac{ - 10}{ - 30} = + \frac{1}{3} = + 0.33$

Since, magnification is positive, we can say that the image is erect and virtual. The size of the image is reduced to one-third in size than the object after refraction.

Explanation:

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