A) 1 : 1
B) 2 : 1
C) \[2\sqrt{2}:1\]
D) 3 : 2
Correct Answer: C
Solution :
Illuminance at A, \[{{I}_{A}}=\frac{L}{{{h}^{2}}}\] Illuminance at B, \[{{I}_{B}}=\frac{L}{\sqrt{{{({{h}^{2}}+{{r}^{2}})}^{2}}}}\cos \theta \] \[=\frac{Lh}{{{({{r}^{2}}+{{h}^{2}})}^{3/2}}}\] \[\therefore \frac{{{I}_{A}}}{{{I}_{B}}}={{\left( 1+\frac{{{r}^{2}}}{{{h}^{2}}} \right)}^{3/2}}\]\[={{\left( 1+\frac{{{8}^{2}}}{{{8}^{2}}} \right)}^{3/2}}={{2}^{3/2}}=2\sqrt{2}:1\]You need to login to perform this action.
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