Question

Glass has refractive index ' \(n\) ' with respect to air and ' \(\theta\) ' is the critical angle for a ray of light going from glass to air. If a ray of light is incident from air on the glass with angle of incidence ' \(\theta\) ', corresponding angle of refraction is

• A. \(\mathrm{sin}^{-1}(\frac{1}{n})\)
• B. \(\mathrm{sin}^{-1}(\mathrm{n})\)
• C. \(\mathrm{sin}^{-1}(\frac{1}{\mathrm{n}^{2}})\)
• D. \(\mathrm{sin}^{-1}(\frac{1}{\sqrt{n}})\)

Hint:

At the critical angle sin theta equals 1 over n; then apply Snell's law for incidence from air at that same angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For glass to air, the critical angle satisfies \(\sin\theta = \dfrac1n\).

Step 2: Apply Snell's Law
For a ray from air entering glass at angle of incidence \(\theta\): \(\sin\theta = n\sin r\).

Step 3: Solve
\(\sin r = \dfrac{\sin\theta}{n} = \dfrac{1}{n}\times\dfrac1n = \dfrac{1}{n^2}\).

Step 4: Result
\(r = \sin^{-1}\left(\dfrac{1}{n^2}\right)\), option C.

Final Answer:
The angle of refraction is sin inverse of 1 over n squared. \[ \boxed{\text{(C) }\sin^{-1}\left(\dfrac{1}{n^{2}}\right)} \]