Question

A satisfactory photographic print is obtained when the exposure time is 10 sec 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 sec
• B. 10 sec
• C. 15 sec
• D. 20 sec

Answer 1

Answer: (D)

20 sec

Answer 2

$I_{1}D_{1}^{2}t_{1} = I_{2}D_{2}^{2}t_{2}$

Here D is constant and $I = \frac{L}{r^{2}}$

So $\frac{L_{1}}{r_{1}^{2}} \times t_{1} = \frac{L_{2}}{r_{2}^{2}} \times t_{2} \Rightarrow \frac{60}{(2)^{2}} \times 10 = \frac{120}{(4)^{2}} \times t \Rightarrow 20\mspace{6mu}\sec$

($I_{1}D_{1}^{2}t_{1} = I_{2}D_{2}^{2}t_{2}$