Question

You can determine the index of refraction of a substance by determining its critical angle.

(a) What is the index of refraction of a substance that has a critical angle of 53.3° when submerged in water, which has an index of refraction of 1.333?

(b) What would the critical angle be for this substance in air?

Answer 1

Answer 2

(a) Index of refraction of the substance (submerged in water):

Using the critical angle formula for refraction from the substance to water:

$$n_{\text{substance}}\sin ({\theta }_c)=n_{\text{water}}\sin (90\mathrm{°})$$$$n_{\text{substance}}\sin (53.3\mathrm{°})=1.333\times 1$$$$n_{\text{substance}}=\frac{1.333}{\sin (53.3\mathrm{°})}=\frac{1.333}{0.802}\approx 1.662$$

Answer:
The index of refraction of the substance is $\boxed{{\displaystyle 1.662}}$.

---

(b) Critical angle for the substance in air:

Using the critical angle formula for refraction from the substance to air:

$$n_{\text{substance}}\sin ({\theta }_c)=n_{\text{air}}\sin (90\mathrm{°})$$$$1.662\sin ({\theta }_c)=1\times 1$$$$\sin ({\theta }_c)=\frac{1}{1.662}\approx 0.602$$$${\theta }_c=\arcsin (0.602)\approx 37.0\mathrm{°}$$

Answer:
The critical angle for the substance in air is $\boxed{{\displaystyle 37.0\mathrm{°}}}$