A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical center of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?

If the object is now moved a little away from the lens, then the screen should be moved towards the lens to get a sharp image of the object on it again.

Magnification of the image decreases or reduces on moving the object away from the lens.

Explanation

Here,

Object distance = $u$ = $-$12 cm

Image distance = $v$ = $+$24 cm

Using the lens formula, we have-

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

Substituting the given values we get-

$\frac{1}{f}=\frac{1}{24}-\frac{1}{(-12)}$

$\frac{1}{f}=\frac{1}{24}+\frac{1}{12}$

$\frac{1}{f}=\frac{1+2}{24}$

$\frac{1}{f}=\frac{3}{24}$

$\frac{1}{f}=\frac{1}{8}$

$f=8cm$

The focal length of the lens is 8cm.

Now if the object is moved away from the lens, the screen has to be moved towards the lens because when the object is moved away from the lens, the object distance is increased. Hence, by the lens formula, the image distance decreases.

Magnification of the lens is given as-

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

Here, we can see that the magnification $m$ is directly proportional to the image distance $v$ and inversely proportional to object distance $u$.

So, on moving the object away from the lens the magnification of the image decreases or reduces, and also the image distance $v$ decreases, on decreasing the value of magnification.

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Updated on: 10-Oct-2022

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