Q. A circular disc is placed in front of a narrow source. When the point of observation is $2\ m$ from the disc, then it covers first HPZ. The intensity at this point is $I$. When the point of observation is $25\ cm$ from the disc then intensity will be
A) $\left( \frac{R_{6}}{R_{2}} \right)^{2}I$
B) $\left( \frac{R_{7}}{R_{2}} \right)^{2}I$
C) $\left( \frac{R_{8}}{R_{2}} \right)^{2}I$
D) $\left( \frac{R_{9}}{R_{2}} \right)^{2}I$
Explanation
(d) $I = \frac{R_{2}^{2}}{4}$ given $n_{1}b_{1} = n_{2}b_{2} \Rightarrow 1 \times 200 = n_{2} \times 25$
$\therefore n_{2} = 8\ HPZ$
$\therefore I = \left( \frac{R_{9}}{2} \right)^{2}$
$= \left( \frac{R_{9}}{R_{8}} \times \frac{R_{8}}{R_{7}} \times \frac{R_{7}}{R_{6}} \times \frac{R_{6}}{R_{5}} \times \frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}} \times \frac{R_{2}}{R_{2}} \right)^{2}$
$= \left( \frac{R_{9}}{R_{2}} \right)^{2}I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.