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

68.4º when submerged in water?

What is the substance, based on Table

25.1? (b) What would the critical

angle be for this substance in air?

Answer 1

Explanations & Calculations

• For the first case,

$\qquad\qquad \begin{aligned} \small n_s.\sin C&=\small n_w.\sin 90\\ \small \frac{n_s}{n_w}&=\small \frac{1}{\sin C}\\ &=\small \frac{1}{\sin 68.4}=1.076\cdots(\text{w.r.t water})\\ \\ \small \frac{n_s}{a_{air}}\times\frac{n_{air}}{n_w}&=\small 1.076\\ \small \frac{n_s}{n_{air}}=n_{substance}&=\small 1.076\times\frac{n_w}{n_{air}}\\ &=\small 1.076\times1.33\\ &=\small 1.43\cdots{\text{w.r.t air}} \end{aligned}$

• You can find what the substance is referring the table for this value.

• When the substance is in air,

$\qquad\qquad \begin{aligned} \small n_s.\sin C&=\small n_{air}.\sin90\\ \small \sin C&=\small \frac{n_{air}.1}{n_s}\\ &=\small \frac{1}{1.43}\\ \small C&=\small \sin^{-1}(1/1.43)\\ &=\small 44.4^0 \end{aligned}$