When an object moves towards a convex lens, the size of the image formed :

decreases

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increases

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first decreases then increases

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remains the same

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Solution

The correct option is B increases
Magnification $= \frac{v}{u}$
and for a convex lens
when $u = - \infty$ $v = f$
$u = - 2 f$ $v = 2 f$
$u = - f$ $v = \infty$
Hence magnification is increasing when an object moves towards a convex lens.

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Similar questions

When an object moves towards a convex lens from beyond 2F upto F, the size of the image _____________.

(a) decreases
(b) increases
(c) remains the same
(d) first decreases then increases

Q. When an object moves closer to a convex lens, always remaining beyond F, the image formed by it shifts _____________.

(a) away from the lens
(b) first away and then towards the lens
(c) towards the lens
(d) first towards and then away from the lens

Q. The distance between an object and a convex lens is changing, and it is noticed that the size of the image formed on a screen is decreasing. Is the object moving in a direction towards the lens or away from it?

Q. A convex lens forms an image of an object on a screen

30 c m

from the lens. When the lens is moved

90 c m

towards the object, then the image is again formed on the screen. Then, the focal length of the lens is

Question 21

The distance between an object and a convex lens is changing. It is noticed that the size of the image formed on a screen is decreasing. Is the object moving in a direction towards the lens or away from it?

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When an object moves towards a convex lens, the size of the image formed :

The correct option is B increasesMagnification =vuand for a convex lens when u=−∞ v=f u=−2f v=2f u=−f v=∞ Hence magnification is increasing when an object moves towards a convex lens.

The correct option is B increases
Magnification $= \frac{v}{u}$
and for a convex lens
when $u = - \infty$ $v = f$
$u = - 2 f$ $v = 2 f$
$u = - f$ $v = \infty$
Hence magnification is increasing when an object moves towards a convex lens.