# The angle $\theta$ of the prism is [closed] A parallel beam of light is incident from air at an angle $a$ on the side $PQ$ of a right angled triangular prism of refractive index $n = \sqrt{2}$. Light undergoes total internal reflection in the prism at the face $PR$ when $a$ has a minimum value of $45°$. The angle $\theta$ of the prism is ...

Snell's law is given by

$$n_1 \sin \theta_1 = n_2 \sin \theta_2$$

However, I'm not sure about the equation and the way we need to follow. Can you brighten me up?

## closed as off-topic by Void, Gert, stafusa, Jon Custer, sammy gerbilNov 24 '17 at 3:55

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• Can you draw the path of the light inside the prism? What is the condition for total internal reflection? – Floris Nov 23 '17 at 19:54
• Yes, I drew it. What do you mean by internal reflection? – Cargobob Nov 23 '17 at 20:34
• The question states "light undergoes total internal reflection at PR". Do you know what that means? – Floris Nov 23 '17 at 20:37
• No, First time I've heard it. I currently don't have any idea about what total internal reflection at $PR$ means. – Cargobob Nov 23 '17 at 20:41
• I have written an answer with the diagram that should help. Review your notes, or google "total internal reflection", if that is still not clear enough. – Floris Nov 23 '17 at 20:48

## 1 Answer

This is the diagram you need: The critical angle $\alpha_c$ is the one at which you just get total internal reflection - that is, the transmitted light would make an angle of 90° to the normal. That means that $\sin\alpha_c = \frac{n_1}{n_2}$.

That should be the hint that gets you going...