Question

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is ___. ($\tan^{-1}(1.52) = 57.7^\circ$, refractive indices of air and glass are 1.00 and 1.52, respectively.)

• A. $39.6^\circ$
• B. $32.3^\circ$
• C. $42.6^\circ$
• D. $36.3^\circ$

Hint:

When reflected light is completely polarized, reflected and refracted rays are at $90^\circ$. Hence $r = 90^\circ - i_p$.

Answer 1

The Correct Option is B

To find the angle of the refracted beam when unpolarized light falls on a glass plate, resulting in a linearly polarized reflected beam, we should consider Brewster's Law. According to Brewster's Law, the angle at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection, is called the Brewster angle.

The law states that when light is incident at the Brewster angle, the reflected light is perfectly polarized perpendicular to the plane of incidence. The relationship between the angle of incidence (\(\theta_B\)), and the refractive indices of the two media (\(n_1\) for air and \(n_2\) for glass) is given by: 

\(\tan(\theta_B) = \frac{n_2}{n_1}\)

In this context:

• \(n_1 = 1.00\) (refractive index of air)
• \(n_2 = 1.52\) (refractive index of glass)

Given that \(\tan^{-1}(1.52) = 57.7^\circ\), this angle (\(\theta_B\)) is the Brewster angle where reflection causes the light to be polarized.

According to Snell's Law, the relationship between the angle of incidence (\(\theta_B\)) and the angle of refraction (\(\theta_r\)) is:

\(n_1 \sin(\theta_B) = n_2 \sin(\theta_r)\)

Substituting values and using the angle we have:

\(1.00 \cdot \sin(57.7^\circ) = 1.52 \cdot \sin(\theta_r)\)

Solving for \(\sin(\theta_r)\):

\(\sin(\theta_r) = \frac{\sin(57.7^\circ)}{1.52}\)

Calculating \(\sin(57.7^\circ)\):

\(\sin(57.7^\circ) \approx 0.8415\)

Then,

\(\sin(\theta_r) = \frac{0.8415}{1.52} \approx 0.5536\)

Finding \(\theta_r\):

\(\theta_r = \sin^{-1}(0.5536) \approx 32.3^\circ\)

Therefore, the angle of the refracted beam with respect to the normal is 32.3°.

Thus, the correct option is $32.3^\circ$.