A converging beam of rays in incident on a diverging lens. Having passed through the lens the rays intersect at a point `15 cm` from the lens. If the lens is removed, the point where the rays meet, move `5 cm` closer to the mounting that holds the lens. Find the focal length of the lens.

Options

A. `-10cm`
B. 20 cm
C. `-30cm`
D. 5 cm

Answer

C

Explanation

By lens formula,
`(1)/(v)-(1)/(u)=(1)/(f)`
`u=10cm`
`v=15cm`
`f=?`
Putting the value, we get
`(1)/(15)-(1)/(10)=(1)/(f)`
`*(10-15)/(150)=(1)/(f)" "therefore f=-(150)/(3)=-30cm`