Optics Question 164
Question: 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 I0. The intensity at a point distance 25 cm from the disc will be
Options:
A) $ I _{1}=0.531I _{0} $
B) $ I _{1}=0.053I _{0} $
C) $ I _{1}=53I _{0} $
D) $ I _{1}=5.03I _{0} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 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}}={{( \frac{R _{5}}{2} )}^{2}}={{( \frac{R _{5}}{R _{4}}\times \frac{R _{4}}{R _{3}}\times \frac{R _{3}}{R _{2}} )}^{2}}\times {{( \frac{R _{2}}{2} )}^{2}} $ or $ 1={{[0.9]}^{6}} $
$ I _{1}=0.531I _{0} $ Hence the correct answer will be (a).
