Question

To get total internal reflection at the interface of water (refractive index 1.33) and a plastic whose refractive index is 1.46: (a) Which material must the light start in? (1) It doesn’t matter, (2) Water, (3) Plastic, (b) What is the critical angle?

Short answer

Therefore, the correct option is (3): plastic

Step-by-step solution

Write the given data from the question.

Refractive index of water,$n_2=1.33$

Refractive index of plastic,$n_1=1.46$

Determine the relation to that relates the refractive index of water and plastic.

The expression that relates the refractive index of two medium: the Snell’s law is given as follows.

${\mathit{n}}_{\mathbf{1}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathit{n}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{2}}$ …… (i)

Here, ${\mathit{\theta }}_{\mathbf{2}}$is the angle of refection and localid="1668593025790" ${\mathit{\theta }}_{\mathbf{1}}$is the angle of incident.

Find out which material must the light start in.

To get the total internal reflection, the must travel from the denser medium to rarer medium and the angle of the incident in the denser medium must be greater than critical angle.

The refractive index of plastic is more than the refractive index of water. Therefore,

$n_1>n_2$

Hence the light starts from the plastic.