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. Then height of this image is 4 cm . The distance between the object and the screen is 90 cm.

Options

A. The distance between the two positions of the lens is 30 cm
B. The distance of the object from the lens in its first position is 36 cm
C. The height of the object is 6 cm
D. The focal length of the lens is 21.6 cm

Answer

A,D

Explanation

Height of object = `sqrt(" H of image" xx H "of image "2) = sqrt(I_(1)I_(2)) = sqrt(9 xx4)`
Height of object = 6 cm
`m_(1) = (I_(1))/(O) = (v_(1))/(u) implies (9)/(6) = (v_(1))/(u_(1))`
`u_(2) = v_(1) = (3)/(2) v_(1) therefore X = (3)/(2)u_(1) - u_(1) = (u_(1))/(2) = X`
` D = 90 = u_(2) + u_(1) = (3)/(2) u_(1) + u_(1)`
`u_(1) = 36 implies X = (36)/(2) = 18`
`therefore` Distance between two position of lens = X = 18
Focal length `f = (D^(2) - V^(2))/(4D) = (90^(2) - 18^(2))/(4 xx 90) = 21.6 "cm"`