Total Internal Reflection — Question 5

Question 5

Immersing a prism can remove its internal reflecting action

A ray enters one leg of a 45-45-90-degree prism normally and strikes its hypotenuse internally at 45 degrees. The prism index is np=1.60. The external medium surrounding the hypotenuse has index ne.

1. Derive the condition on ne for total internal reflection.

2. Classify the behavior in air with ne=1.00 and in a liquid with ne=1.33. For the latter, find the refracted angle.

Question 5 - Solution

Immersing a prism can remove its internal reflecting action

(a) Method and calculation

$$\sin45^\circ>n_e/n_p\Rightarrow\boxed{n_e<n_p/\sqrt2}.$$$$n_p/\sqrt2=1.13137\Rightarrow\boxed{\text{air: TIR; liquid: no TIR}}.$$

(b) Interpretation and further calculation

$$n_p\sin45^\circ=n_e\sin\theta_t.$$$$\boxed{\theta_t=\sin^{-1}\!\left(\frac{1.60}{1.33\sqrt2}\right)=58.2830^\circ}.$$

Physical check

The surrounding index matters at the reflecting face even though the ray entered the first face normally.