A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected by upper surface and 50 % of light is reflected from lower surface. Rest of the light is refracted. Abo

Question

A parallel beam of light of intensity I is incident on a glass plate. 25 % of light is reflected by upper surface and 50 % of light is reflected from lower surface. Rest of the light is refracted. Above two reflected rays interfere, the ratio of maximum to minimum intensity in interference region is

TestHub · Accepted Answer

Let the intensity of the incident light be $I$ .

Intensity of light reflected from the upper surface is:
$I_{1} = 25% \times I = \frac{I}{4}$

Intensity of light refracted into the glass plate is:
$I_{t} = I - \frac{I}{4} = \frac{3 I}{4}$

At the lower surface, $50%$ of this light is reflected. Thus, the intensity of light reflected from the lower surface is:
$I_{2} = 50% \times \frac{3 I}{4} = \frac{3 I}{8}$

This light emerges from the upper surface to interfere with the first reflected ray. Neglecting any further reflection at the upper boundary, the intensities of the two interfering rays are $I_{1} = \frac{I}{4}$ and $I_{2} = \frac{3 I}{8}$ .

The ratio of maximum to minimum intensity in the interference region is:
$\frac{I _{ma x}}{I _{min}} = (\frac{I _{1} + I _{2}}{I _{1} - I _{2}})^{2}$

Substituting the values of $I_{1}$ and $I_{2}$ :
$\frac{I _{ma x}}{I _{min}} = \frac{\frac{I}{4} + \frac{3 I}{8}}{\frac{I}{4} - \frac{3 I}{8}}^{2} = \frac{\frac{1}{2} + \frac{3}{8}}{\frac{1}{2} - \frac{3}{8}}^{2}$

Answer: $\frac{\frac{1}{2} + \frac{3}{8}}{\frac{1}{2} - \frac{3}{8}}^{2}$

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