Question

A beam of light both reflects and refracts at the surface between air and glass. The index of refraction of the glass is \(1.4\). If the refracted and the reflected rays are perpendicular to each other, then the angle of incidence in the air is

• A. \(\tan^{-1}(1.4)\)
• B. \(\sin^{-1}\left(\dfrac{1}{1.4}\right)\)
• C. \(\tan^{-1}\left(\dfrac{1}{1.4}\right)\)
• D. \(\sin^{-1}\left(\dfrac{1.4}{\pi}\right)\)

Hint:

When the reflected ray and refracted ray are perpendicular to each other, the angle of incidence is called Brewster’s angle and is given by \[ \tan i=\mu \]

Answer 1

The Correct Option is A

Step 1: State the given condition.
When the reflected and refracted rays are perpendicular to each other, the angle between them is $90^\circ$.

Step 2: Use geometry to relate the angles.
Let $i$ be the angle of incidence and $r$ be the angle of refraction. The reflected ray makes angle $i$ with the normal on the same side.
The angle between the reflected ray and the refracted ray is $(90^\circ - i) + (90^\circ - r) = 180^\circ - i - r$. Wait - more simply: reflected ray is on the same side as incident, making angle $i$ with the normal. Refracted ray makes angle $r$ on the other side. The angle between reflected and refracted rays is $180^\circ - i - r$. Setting this to $90^\circ$ gives:
\[ i + r = 90^\circ \implies r = 90^\circ - i \]

Step 3: Apply Snell's law at the air-glass interface.
\[ n_1 \sin i = n_2 \sin r \] With $n_1 = 1$ (air) and $n_2 = 1.4$ (glass):
\[ \sin i = 1.4 \sin(90^\circ - i) = 1.4 \cos i \]

Step 4: Divide both sides by $\cos i$.
\[ \frac{\sin i}{\cos i} = 1.4 \implies \tan i = 1.4 \]

Step 5: Solve for the angle of incidence.
\[ i = \tan^{-1}(1.4) \]

Step 6: State the answer.
This angle is known as Brewster's angle. At this angle, the reflected light is completely polarised.
\[ \boxed{\tan^{-1}(1.4)} \]