A) 3.1 sec
B) 1 sec
C) 12.8 sec
D) 16 sec
Correct Answer: C
Solution :
To develop a print a fix amount of energy is required. Total light energy incident on photo print \[I\times A\,t=\frac{L}{{{r}^{2}}}A\,t\]\[\Rightarrow \frac{{{L}_{1}}}{r_{1}^{2}}{{A}_{1}}{{t}_{1}}=\frac{{{L}_{2}}}{r_{2}^{2}}{{A}_{2}}{{t}_{2}}\] \[\Rightarrow \frac{{{t}_{1}}}{r_{1}^{2}}=\frac{{{t}_{2}}}{r_{2}^{2}}\] (\[\because \ {{L}_{1}}={{L}_{2}}\] and \[{{A}_{1}}={{A}_{2}}\]) \[\Rightarrow {{t}_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}.{{t}_{1}}\]\[=\left( \frac{0.40}{0.25} \right)\ 2\times 5\]= 12.8 sec.
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