question_answer

A source of light suspended above a circular table at a height equal to the radius of the table gives an intensity I at the centre of the table, the intensity at the edge of the table would be (Assuming illuminace remains the same)

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 \[\approx \] 2.87