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A ray of light of intensity I is incident on a parallel glass slab at point A as shown in diagram. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A' B' undergo interference. The ratio of I_{max} and I_{min} is :

 

Option: 1

49:1
 


Option: 2

7:1


Option: 3

4:1

 


Option: 4

8:1


Answers (1)

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\begin{aligned} &\text { From figure } I_{1}=\frac{I}{4} \text { and } I_{2}=\frac{9 I}{64} \Rightarrow \frac{I_{2}}{I_{1}}=\frac{9}{16}\\ & \text{By using} \ \ \frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{\frac{I_{2}}{I_{1}}} 1}{\sqrt{\frac{I_{2}}{I_{1}}}-1}\right)=\left(\frac{\sqrt{\frac{9}{16}+1}}{\sqrt{\frac{9}{16}}-1}\right)=\frac{49}{1} \end{aligned}

 

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avinash.dongre

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