When exposed to a 10 W source for 12 s , a photographic plate...

The intensity of the lines on a photographic plate is determined by the total light energy received, which is the product of the power of the source and the exposure time.

For equally intense lines, the total energy must be the same in both cases.

$E_1 = E_2$

$P_1 t_1 = P_2 t_2$

Given $P_1 = 10 \text{ W}$, $t_1 = 12 \text{ s}$, and $P_2 = 12 \text{ W}$.

$10 \times 12 = 12 \times t_2$

$t_2 = \dfrac{120}{12} = 10 \text{ s}$