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 $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक