A) 0.251
B) 0.51
C) 0.77
D) 2.87
Correct Answer: D
Solution :
Let A be the source of light, then illuminance (E) is given by \[E=\frac{I\cos \theta }{{{(\sqrt{2r})}^{2}}}=\frac{I}{2{{r}^{2}}}\times \cos \theta \] where \[\theta ={{45}^{o}}\], \[\therefore \] \[\cos {{45}^{o}}=\frac{1}{\sqrt{2}}\] \[\therefore \] \[E=\frac{I}{2{{r}^{2}}}\times \frac{1}{\sqrt{2}}\] ... (i) Also, illuminance (E) at centre of table is \[E=\frac{I}{2{{r}^{2}}}\] ... (ii) Given, E = E \[\therefore \] \[\frac{I}{2\sqrt{2}\,{{r}^{2}}}=\frac{I}{{{r}^{2}}}\] \[\Rightarrow \] \[I=2\sqrt{2}\,I\] = 2.828 I \[\approx \] 2.87You need to login to perform this action.
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