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

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 $A^{1} B^{1}$ undergo interference. The ratio $I_{max} / I_{min}$ is

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Let the intensity of the incident ray at point A be $I$ .

At point A, the ray undergoes partial reflection and refraction. Since $25%$ of the incident energy is reflected, the intensity of the reflected ray $A B$ is:
$I_{1} = 25% of I = \frac{I}{4}$

The remaining energy is transmitted, so the intensity of the refracted ray $A C$ is:
$I_{A C} = 75% of I = \frac{3 I}{4}$

At point C, the ray $A C$ is reflected. The intensity of the reflected ray $C A^{1}$ is $25%$ of $I_{A C}$ :
$I_{C A^{1}} = \frac{1}{4} \times \frac{3 I}{4} = \frac{3 I}{16}$

At point $A^{1}$ , the ray $C A^{1}$ is refracted out of the glass slab. The transmitted energy is $75%$ of the incident energy at $A^{1}$ . Thus, the intensity of the emergent ray $A^{1} B^{1}$ is:
$I_{2} = \frac{3}{4} \times I_{C A^{1}} = \frac{3}{4} \times \frac{3 I}{16} = \frac{9 I}{64}$

The ratio of the amplitudes of the interfering rays $A B$ and $A^{1} B^{1}$ is:
$\frac{a _{1}}{a _{2}} = \frac{I _{1}}{I _{2}} = \frac{I /4}{9 I /64} = \frac{16}{9} = \frac{4}{3}$

The ratio of maximum to minimum intensity in the interference pattern is given by:
$\frac{I _{max}}{I _{min}} = (\frac{a _{1} + a _{2}}{a _{1} - a _{2}})^{2}$

Substituting the ratio of amplitudes:
$\frac{I _{max}}{I _{min}} = (\frac{4 + 3}{4 - 3})^{2} = (\frac{7}{1})^{2} = \frac{49}{1}$

Therefore, the ratio $I_{max} / I_{min}$ is $49 : 1$ .

Answer: $49 : 1$

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