The diagram alongside shows the refraction of a ray of light from air to a liquid.

(a) Write the values of (i) angle of incidence, and (ii) angle of refraction.

(b) Use Snell's law to find the refractive index of liquid with respect to air.

(a) (i) The angle of incidence is the angle which the incident ray makes with the normal. Hence, ∠i = 90° – 30° = 60°.

(ii) Angle of refraction is the angle which the refracted ray makes with the normal Hence, ∠r = 90° – 45° = 45°.

(b) We know that according to Snell’s law,

$_{\text{air}}\text{μ}_{\text{liquid}} = \dfrac{\text{sin i}}{\text{sin r}} \\[0.5em] = \dfrac{\text{sin 60} \degree}{\text{sin 45} \degree} \\[0.5em] = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} \\[0.5em] = \dfrac{\sqrt{3}}{\sqrt{2}} \\[0.5em] = 1.22$

Hence, the refractive index of liquid with respect to air is 1.22