Question

The Brewster's angle for the air-glass interface is \((54.74)^{\circ }\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ }\), then the angle of refraction is
\(\left(\tan (54.74)^{\circ }=\sqrt{2},\,\sin 45=1/\sqrt{2}\right)\)

• A. \(\sin ^{-1}(\sqrt{2})\)
• B. \(\sin ^{-1}\left(\frac{1}{2}\right)\)
• C. \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
• D. \(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

Hint:

Brewster's law gives the refractive index; then apply Snell's law at 45 degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Brewster's law: \(\mu=\tan i_B=\tan(54.74^{\circ})=\sqrt{2}\).

Step 2: Snell's law
\(\sin45^{\circ}=\mu\sin r\)
\(\dfrac{1}{\sqrt{2}}=\sqrt{2}\sin r\)

Step 3: Solve
\(\sin r=\dfrac{1}{2}\), so \(r=\sin^{-1}\left(\dfrac{1}{2}\right)\) (which is \(30^{\circ}\)).

Final Answer:
The angle of refraction is sin inverse of one half. \[ \boxed{\text{(B) }\sin^{-1}\left(\dfrac{1}{2}\right)} \]