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 size of the object to size of shorter image is 4.
B. Distance between two position of lens is 32 cm
C. focal length of lens is `(64)/(3)` cm
D. When the shorter image is formed on screen,distance of lens from the screen is 32 cm.

Answer

B, C, D

Explanation

`m_(1)=(v)/(u),m_(2)=(u)/(v)`
Ratio of size of two images
`(m_(1))/(m_(2))=((v)/(mu))/((u)/(v))=(v^(2))/(v^(2))/(u^(2))`
`(v^(2))/(u^(2))=4` .(i)
`impliesv+u=96` .(ii)
(1) & (2) `v=64`
`u=32cm`
ratio of size of object to size of shorter
`=(64)/(32)=2`
Distance between two position
`=v-u=32cm`
`(1)/(f)=(1)/(f)-(1)/(u)=(1)/(64)-(1)/(-32)impliesf=(64)/(3)cm`