Question

A photographic plate placed at a distance of $5cm$ from a weak point source is exposed for $3s$. if the plate is kept at a distance of $10cm$ from the source, the time needed for the same exposure is

• A. $3s$
• B. $12s$
• C. $24s$
• D. $48s$

Answer 1

The correct option is B $12s$

The intensity is the power delivered per unit area, and hence it is inversely proportional to the square of the distance from source.

$I=\frac{k}{d^2}$

initially,

$I_0=\frac{k}{25}$

Exposure $=3I_0=\frac{3k}{25}$

Then distance is changed to 10cm

$I_2=\frac{k}{100}$

Time to get same exposure = t.

$\frac{kt}{100}=\frac{3k}{25}$

We get $t=12s$
So, the answer is option (B).