Q. A circular disc is placed in front of a narrow source. When the point of observation is at a distance of $1\ meter$ from the disc, then the disc covers first HPZ. The intensity at this point is $I_{0}$. The intensity at a point distance $25\ cm$ from the disc will be (If ratio of consecutive amplitude of HPZ is 0.9)
A) $I_{1} = 0.531I_{0}$
B) $I_{1} = 0.053I_{0}$
C) $I_{1} = 53I_{0}$
D) $I_{1} = 5.03I_{0}$
Explanation
(a) $I_{0} = R^{2} = \frac{R_{2}^{2}}{4}$
Number of $HPZ$ covered by the disc at $b = 25cm$
$n_{1}b_{1} = n_{2}b_{2}$
$n_{2} = \frac{n_{1}b_{1}}{b_{2}} = \frac{1 \times 1}{0.25} = 4$
Hence the intensity at this point is
$I = {R'}^{2} = \left( \frac{R_{5}}{2} \right)^{2} = \left( \frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}} \right)^{2} \times \left( \frac{R_{2}}{2} \right)^{2}$
$I = (0.9)^{6}I_{0}$
$I_{1} = 0.531I_{0}$
Hence the correct answer will be (a)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.