A) 40 %
B) 64%
C) 78%
D) 92%
Correct Answer: C
Solution :
\[{{I}_{1}}=\frac{L}{r_{1}^{2}}=\frac{L}{16}\] and \[{{I}_{2}}=\frac{L}{r_{2}^{2}}=\frac{L}{9}\] % increase in illuminance \[=\frac{{{I}_{2}}-{{I}_{1}}}{I}\times 100\]\[=\left( \frac{16}{9}-1 \right)\times 100\]\[\approx 78\,%\]
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