Q. A satisfactory photographic print is obtained when the exposure time is $10\ s$ at a distance of $2\ m$ from a 60 $cd$ lamp. The time of exposure required for the same quality print at a distance of $4\ m$ from a 120 $cd$ lamp is
A) $5\ s$
B) $10\ s$
C) $15\ s$
D) $20\ s$
Explanation
(d) $I_{1}D_{1}^{2}t_{1} = I_{2}D_{2}^{2}t_{2}\ \lbrack D$ is diameter of aperature]
Here $D$ is constant and $I = \frac{L}{r^{2}}$
So $\frac{L_{1}}{r_{1}^{2}} \times t_{1} = \frac{L_{2}}{r_{2}^{2}} \times t_{2} \Rightarrow \frac{60}{(2)^{2}} \times 10 = \frac{120}{(4)^{2}} \times t \Rightarrow 20\ s = t$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.