Question 261: Wave Optics

Question

A ray of light of intensity ' I ' is incident on a parallel glass slab at a point ' $A$ ' as shown in figure. It undergoes partial reflection and refraction. At each reflection $25 \%$ of incident energy is reflected. The rays $A B$ and $A B$ undergo interference. The ratio $\frac{\mathrm{I}_{	ext {max }}}{\mathrm{I}_{	ext {min }}}$ is
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1meu8s8ww/b2538cce-7791-4c19-b453-8d394ad882a7/4cf5e200-836a-11f0-9311-29d163b3c052/file-jaoe38c1meu8s8wx.png?format=png)

Accepted Answer

Given that, $25 \%$ of the incident light is reflected at the upper surface.
If incident intensity is $\mathrm{I}_0$, then
Reflected at A (upper surface) $\mathrm{I}_1=25 \mathrm{I}$
Transmitted $=0.75 \mathrm{I}$
At lower surface:
$25 \%$ of this is reflected back:
$ I_2=\frac{1}{4} 	imes \frac{3}{4} I_0=\frac{3}{16} I $
Again at upper surface:
$ \begin{aligned} & 	ext { Reflected }=\frac{1}{4} 	imes \frac{3}{16} I=\frac{3}{64} I \ & 	ext { Trasmitted }=\frac{3}{16} I-\frac{3}{64} I=\frac{9}{64} I \end{aligned} $
$ \begin{aligned} &\because \quad 	ext { Intensity } \propto 	ext { (Amplitude) }{ }^2\ &\begin{aligned} \Rightarrow \quad A_1 & =\sqrt{\frac{I}{4}}=\frac{\sqrt{I}}{2}, A_2=\sqrt{\frac{9}{64} I}=\frac{3}{8} \sqrt{I} \ \frac{I_{\max }}{I_{\min }} & =\left(\frac{A_1+A_2}{A_1-A_2}ight)^2 \ & =\left(\frac{\frac{\sqrt{I}}{2}+\frac{3}{8} \sqrt{I}}{\frac{\sqrt{I}}{2}-\frac{3}{8} \sqrt{I}}ight)^2=\left(\frac{\frac{7}{8} \sqrt{I}}{\frac{1}{8} \sqrt{I}}ight)^2=\frac{49}{1} \end{aligned} \end{aligned} $

Answer

7:1

Answer

$49: 1$

Answer

4:1

Answer

$8: 1$

Wave Optics

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