In displacement method, the distance between object and screen is 96cm. The ratio of lengths of two images formed by a converging lens placed between them is 4. Then,

Options

A. ratio of the length of object to the length of shorter image is 2
B. distance between the two positions of the lens is 32 cm
C. focal length of the lens is `64//3` cm
D. when the shorter image is formed on screen, distance of the lens from the screen is 32cm.

Answer

a.,b.,c.,d.

Explanation

`(h_(1))/(h_(2))=4`
`(h_(1))/(h_(2))xx(h_(2))/(h_(0))=1rArr (h_(2))/(h_(0))=0`
When shorter image is formed, magnification` = 2`
`rArr |(v)/(u)|=2` and `|v|+|u|=96`
`rArr |v|=64` and `|u|=32`
Hence, object distance ` =32` and second lens position can be `96-32=64cm`
Distance between two positions `=32cm`
`(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(64)-(1)/((-32))=(1)/(f)rArrf=(64)/(3)cm`