Question

To find the image distance for varying object distances in case of a convex lens of focal length 15 cm, a student obtains on a screen a sharp image of a bright object by placing it at 20 cm distance from the lens. After that he gradually moves the object away from the lens and each time focuses the image on the screen. (a) In which direction-towards or away from the lens does he move the screen to focus the object ? (b) How does the size of image change? (c) Approximately at what distance does he obtain the image of magnification —1 ? (d) How does the intensity of image change as the object moves farther and farther away from the lens?

Answer 1

(a)

The given lens is a convex lens. On moving the screen away from the lens, then the lens, we can focus the object.

(b)

When we move away the size of the object decreases.

(c)

Magnification is ratio of size of object to size of image. The magnification is equal to 1 only which size of object and image is same.

It is same at twice the focal length which is 2F.

From question, focal length is 15 cm.

So, 2F = 2 × 15 = 30 cm

The magnification is 1 at 30 cm.

(d)

When the image is move farther away and away, the image becomes sharper and sharper.