The height of the image formed by a converging lens on a screen is 8 cm. For the same position of the object and screen again, an image of size 12.5 cm is formed on the screen by shifting the lens. The height of the object is (in cm) -

Open in App

Solution

From displacement method
$h_{0} = \sqrt{h I_{I} \times h I_{2}} = \sqrt{8 \times 12.5} = \sqrt{100} = 10 c m$

Suggest Corrections

Similar questions

Q. A

16 c m

long image of an object is formed by a convex lens on a screen. On moving the lens toward the screen. a

9 c m

long image is formed again on the screen. The size of the object is

Q. A 9 cm high image of an object is formed on a screen by a convex lens. When the lens is displaced towards the screen, again a 4 cm high image is formed on the screen. The height of the object is-

Q. A convex lens forms a real image on a screen when an object is placed at 10 cm from the screen. On shifting the lens by 2 cm, a real image is again formed on the screen. Find the focal length of the lens.

Q. A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90cm.

Join BYJU'S Learning Program

The height of the image formed by a converging lens on a screen is 8 cm. For the same position of the object and screen again, an image of size 12.5 cm is formed on the screen by shifting the lens. The height of the object is (in cm) -

From displacement method h0=√hII×hI2=√8×12.5=√100=10 cm

From displacement method
$h_{0} = \sqrt{h I_{I} \times h I_{2}} = \sqrt{8 \times 12.5} = \sqrt{100} = 10 c m$