A) 56%
B) 44%
C) 36%
D) 64%
Correct Answer: C
Solution :
The actual luminous intensity of the lamp is \[{{I}_{1}}\] whereas the intensity is \[{{I}_{1}}^{\prime }\]in the dirty state. I position, \[\frac{{{I}_{1}}^{\prime }}{{{I}_{2}}}={{\left( \frac{x}{10} \right)}^{2}}\] II position, \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{x}{8} \right)}^{2}}\] \[\Rightarrow \frac{{{I}_{1}}^{\prime }}{{{I}_{1}}}=0.64\] \[\Rightarrow \ {{I}_{1}}^{\prime }=0.64\,{{I}_{1}}\]. Thus, % of light absorbed = 36%.You need to login to perform this action.
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