Question

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 $A B$ and $A^{\prime} B^{\prime}$ undergo interference. The ratio of $I_{\max }$ and $I_{\min }$ is :

• A. 49:01:00
• B. 7:01
• C. 4:01
• D. 8:01

Answer 1

The Correct Option is A

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

When a light beam moves from an optically denser to an optically rarer medium, a process known as total internal reflection takes place.

The departure angle of the light is larger than the incident angle when light is incident on a material with a lower refractive index because the beam of light is bent away from the normal. "Internal reflection" is the term used to describe this reflection.

If the incidence angle is larger than the critical angle, entire internal reflection will occur, and the exit angle will then begin to approach 90°.

The angle of refraction is larger than the angle of incidence when light enters from a denser to a rarer medium because it bends away from the normal.

The majority of the light is refracted when the angle of incidence is smaller since both reflection and refraction take place.

As a result, the pictures produced are brighter than those produced by lenses or mirrors.

This is due to the fact that light intensity is always reduced when it is refracted via a lens or reflected off a mirror.

The incident beam of light is reflected back to the medium when the critical angle is lower than the angle of incidence.

The critical angle is the incidence angle that results in a 90 degree refraction angle. Critical angle, which is an angle of incidence value, is 48.6 degrees for the water-to-air limit. The essential angle for the glass-to-crown water barrier is 61.0 degrees. The combination of materials present on each side of the boundary determines the actual value of the critical angle.

If µ_d is the refractive index of the denser medium, from Snell’s Law, the refractive index of air with respect to the denser medium is given by,

µ_a/µ_d = sini/sinr