Question

The refractive index of glass is 1.5 and that of water is 1.33. The critical angle for a ray of light going from glass to water is

• A. $\sin^{-1}\left(\frac{4}{7}\right)$
• B. $\sin^{-1}\left(\frac{5}{8}\right)$
• C. $\sin^{-1}\left(\frac{8}{9}\right)$
• D. $\sin^{-1}\left(\frac{2}{3}\right)$

Hint:

The sine of an angle can never exceed 1. When setting up your critical angle fraction, always place the smaller refractive index in the numerator: $\sin\theta_c = \frac{\mu_{\text{small}}}{\mu_{\text{large}}}$. This rule guarantees you will not accidentally invert the fraction.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question gives the absolute refractive indices of glass ($\mu_g = 1.5$) and water ($\mu_w = 1.33$).
We need to find the critical angle $\theta_c$ for a ray of light traveling from glass (optically denser medium) to water (optically rarer medium).

Step 2: Key Formula or Approach:
The critical angle $\theta_c$ for a light ray traveling from a denser medium ($d$) to a rarer medium ($r$) is given by: $$\sin \theta_c = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}}$$ Here, the denser medium is glass and the rarer medium is water.

Step 3: Detailed Explanation:
Convert the decimal refractive indices into standard fractional forms to simplify calculations: $$\mu_g = 1.5 = \frac{3}{2}$$ $$\mu_w = 1.33 \approx \frac{4}{3}$$ Substitute these values into the critical angle formula: $$\sin \theta_c = \frac{\mu_w}{\mu_g} = \frac{\frac{4}{3}}{\frac{3}{2}}$$ Simplify the compound fraction by multiplying the top numerator by the reciprocal of the bottom denominator: $$\sin \theta_c = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$$ Take the inverse sine of both sides to isolate the critical angle $\theta_c$: $$\theta_c = \sin^{-1}\left(\frac{8}{9}\right)$$

Step 4: Final Answer:
The critical angle is $\sin^{-1}\left(\frac{8}{9}\right)$, which matches option (C).