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Q. A ray of light of intensity $I$ is incident on a parallel glass-slab at the point $A$ and it undergoes partial reflection and refraction as shown in the figure. At each reflection, $25\% \, $ of the incident energy is reflected and the rest is transmitted. If the rays $AB$ and $A′B′ \, $ undergo interference, then the ratio $\frac{I_{max}}{I_{min}}$ is

Question

NTA AbhyasNTA Abhyas 2020Ray Optics and Optical Instruments

Solution:

From figure $I_{1}=\frac{I}{4}$ and $I_{2}=\frac{9 I}{64}$
$\Rightarrow \frac{I_{2}}{I_{1}}=\frac{9}{16}$

Solution
By using $\frac{I_{m a x}}{I_{m i n}}=\left(\frac{\sqrt{\frac{\left(I \right)_{2}}{\left(I ⁡\right)_{1}}} + 1}{\sqrt{\frac{\left(I ⁡\right)_{2}}{\left(I ⁡\right)_{1}}} - 1}\right)^{2}=\left(\frac{\sqrt{\frac{9}{1 6}} + 1}{\sqrt{\frac{9}{1 6}} - 1}\right)^{2}=\frac{4 9}{1}$