Optics Question 165
Question: A wavefront presents one, two and three HPZ at points A, B and C respectively. If the ratio of consecutive amplitudes of HPZ is 4 : 3, then the ratio of resultant intensities at these point will be
Options:
A) 169 : 16 : 256
B) 256 : 16 : 169
C) 256 : 16 : 196
D) 256 : 196 : 16
Show Answer
Answer:
Correct Answer: B
Solution:
$ I _{A}=R _{1}^{2} $
$ I _{B}={{(R _{1}-R _{2})}^{2}}=R _{1}^{2}{{( 1-\frac{R _{2}}{R _{1}} )}^{2}}=R _{1}^{2}{{( 1-\frac{3}{4} )}^{2}}=\frac{R _{1}^{2}}{16} $
$ I _{C}={{(R _{1}-R _{2}+R _{3})}^{2}} $
$ =R _{1}^{2}{{( 1-\frac{R _{2}}{R _{1}}+\frac{R _{3}}{R _{1}} )}^{2}} $
$ =R _{1}^{2}{{( 1-\frac{R _{2}}{R _{1}}+\frac{R _{3}}{R _{2}}\times \frac{R _{2}}{R _{1}} )}^{2}} $
$ =R _{1}^{2}{{( 1-\frac{3}{4}+\frac{3}{4}\times \frac{3}{4} )}^{2}} $
$ ={{( \frac{13}{16} )}^{2}}R _{1}^{2}=\frac{169}{256}R _{1}^{2} $
$ \therefore I _{A}:I _{B}:I _{C}=R _{1}^{2}:\frac{R _{1}^{2}}{16}:\frac{169}{256}R _{1}^{2}=256:16:169 $
BETA
