Q. An electric lamp is fixed at the ceiling of a circular tunnel as shown is figure. What is the ratio the intensities of light at base $A$ and a point $B$ on the wall
A) $1\ :2$
B) $2\ :\sqrt{3}$
C) $\sqrt{3}\ :1$
D) $1\ :\sqrt{2}$
Explanation
(d) $I_{A} = \frac{L}{(2r)^{2}}$ and $I_{B} = \frac{L}{\left( r\sqrt{2} \right)^{2}}\cos\theta$
$= \frac{L}{2r^{2}}.\frac{r}{r\sqrt{2}} = \frac{L}{2\sqrt{2}\ r^{2}}$
$\therefore\frac{I_{A}}{I_{B}} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.