Question

To prepare a print the time taken is 5 sec due to lamp of 60 watt at 0.25 m distance. If the distance is increased to 40 cm then what is the time taken to prepare the similar print

• A. 3.1 sec
• B. 1 sec
• C. 12.8 sec
• D. 16 sec

Answer 1

Answer: (C)

12.8 sec

Answer 2

To develop a print a fix amount of energy is required. Total light energy incident on photo print

$I \times At = \frac{L}{r^{2}}At \Rightarrow \frac{L_{1}}{r_{1}^{2}}A_{1}t_{1} = \frac{L_{2}}{r_{2}^{2}}A_{2}t_{2}$

$\Rightarrow \frac{t_{1}}{r_{1}^{2}} = \frac{t_{2}}{r_{2}^{2}}$ ($\because\mspace{6mu} L_{1} = L_{2}$ and $A_{1} = A_{2}$)

$\Rightarrow t_{2} = \frac{r_{2}^{2}}{r_{1}^{2}}.t_{1} = \left( \frac{0.40}{0.25} \right)\mspace{6mu} 2 \times 5$= 12.8 sec.