If the critical angle for a glass–air boundary is 45º, what is the refractive index of the glass?

Analysis

The question asks for the refractive index of glass, given the critical angle at a glass-air boundary. This is a classic optics problem involving Snell's Law as it relates to total internal reflection.

Concept

The critical angle ($c$) is the angle of incidence in a denser medium for which the angle of refraction in the less dense medium is 90º. The relationship between the refractive index ($n$) and the critical angle ($c$) is given by the formula:

Where:

• $n$ is the refractive index of the denser medium (glass).
• $c$ is the critical angle.

Solution

1. Identify the given values: • Critical angle ($c$) = 45º

2. Apply the formula: $$n = \frac{1}{\sin(45^\circ)}$$

3. Recall the trigonometric value: We know from standard trigonometric ratios that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ (or $\frac{\sqrt{2}}{2}$).

4. Calculate $n$: $$n = \frac{1}{\frac{1}{\sqrt{2}}}$$ $$n = \sqrt{2}$$

Conclusion

The refractive index of the glass is $\sqrt{2}$. This matches option A.

Exam Tip

Always remember that the refractive index ($n$) is always greater than 1. If you ever calculate a value less than 1, check your math! Also, for common angles like 30º, 45º, and 60º, memorizing their sine values will save you precious time during your JAMB exam.