Q. A lamp rated at $100\ cd$ hangs over the middle of a round table with diameter $3\ m$ at a height of $2\ m$. It is replaced by a lamp of $25\ cd$ and the distance to the table is changed so that the illumination at the centre of the table remains as before. The illumination at edge of the table becomes $X$ times the original. Then $X$ is
A) $1/3$
B) $16/27$
C) $1/4$
D) $1/9$
Explanation
(a) Case I
$I_{A} = \frac{100}{2^{2}} = 25\ cd$
and $I_{B} = \frac{100}{(2.5)^{2}}\cos\theta$
$= \frac{100}{{2.5}^{2}} \times \frac{2}{2.5} = \frac{200}{(2.5)^{3}}$
Case II
$I_{B}^{''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''} = X\ I_{B} = \frac{25}{(3.25)^{3/2}}$
so $\frac{I_{B}^{''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''}}{I_{B}} = \frac{25}{200} \times \frac{(2.5)^{3}}{(3.25)^{3/2}}$
$\Rightarrow X = 1/3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.