A satisfactory photographic print is obtained when the exposure time is `10 sec` at a distance of `2m` from a `60 cd` lamp. The time of exposure required for the same quality print at a distance of `4m` from a `120 cd` lamp is

Options

A. `5 sec`
B. `10 sec`
C. `15 sec`
D. `20sec`

Answer

D

Explanation

`I_(1)D_(1)^(2)t_(1)=I_(2)D_(2)^(2)t_(2)`
Here `D` is constant and `I=(L)/(r^(2))`
So `(L_(1))/(r_(1)^(2))xxt_(1)=(L_(2))/(r_(2)^(2))xxt_(2)`
`implies(60)/((2)^(2))xx10=(120)/((4)^(2))xxt=20sec`