Question

In the figure, light enters a $90 °$ triangular prism at point $P$ with incident angle $θ,$ and then some of it refracts at point $Q$ with an angle of refraction of $90 ° .$ (a) What is the index of refraction of the prism in terms of $θ ?$ (b) What numerically is the maximum value that the index of refraction can have? Does light emerge at $Q$ if the incident angle at $P$ is, (c) increased slightly and (d) decreased slightly? (e) What is the incident angle if the index of refraction is $1.30 ?$

Accepted Answer

(a) A ray diagram is shown below:

Let $θ_{1}$ be the angle of incidence and $θ_{2}$ be the angle of refraction at the first surface. Let $θ_{3}$ be the angle of incidence at the second surface. The angle of refraction there is $θ_{4} = 90 ° .$ The law of refraction applied to the second surface that yields $n \sin θ_{3} = \sin θ_{4} = 1 .$ As shown in the diagram, the normals to the surfaces at $P$ and $Q$ are perpendicular to each other. The interior angles of the triangle formed by the ray and the two normals must sum to $180 °,$ so $θ_{3} = 90 ° - θ_{2}$ and $\sin θ_{3} = \sin b 90 ° - θ_{2} g = \cos θ_{2} = \sqrt{1 - \sin^{2} θ_{2}} .$

According to the law of refraction, applied at $Q, n \sqrt{1 - \sin^{2} θ_{2}} = 1 .$ The law of refraction, applied to point $P,$ yields $\sin θ_{1} = n \sin θ_{2},$ so $\sin θ_{2} = \frac{(\sin θ_{1})}{n}$ and

$n \sqrt{1 - \frac{\sin^{2} θ_{1}}{n^{2}}} = 1 .$

Squaring both sides and solving for $n,$ we get

$n = \sqrt{1 + \sin^{2} θ_{1}}$

(b) The greatest possible value of $\sin^{2} θ_{1}$ is $1,$ so the greatest possible value of $n$ is $n_{\max} = \sqrt{2} = 1.41$

(c) For a given value of $n,$ if the angle of incidence at the first surface is greater than $θ_{1}$ , the angle of refraction there is greater than $θ_{2}$ and the angle of incidence at the second face is less than $θ_{3} (= 90 ° - θ_{2})$ . That is, it is less than the critical angle for total internal reflection, so the light leaves the second surface and emerges into the air.

(d) If the angle of incidence at the first surface is less than $θ_{1},$ the angle of refraction there is less than $θ_{2}$ and the angle of incidence at the second surface is greater than $θ_{3} .$ This is greater than the critical angle for total internal reflection, so all the light is reflected at $Q$ .

(e) Let $i$ is the angle of incident at point $P$ ,

From above figure, at point of emergence,

$\frac{\sin r_{2}}{\sin 90 °} = \frac{1}{μ} \sin r_{2} = \frac{1}{μ} \dots (1)$

At point of incident,

$r_{1} = A - r_{2}$

Applying Snell's law,

$1 \times \sin θ = μ \sin r_{1}$

$\sin θ = μ \sin (A - r_{2}) \sin θ = μ (\sin A \cos r_{2} - \sin r_{2} \cos A) (∵ A = \frac{π}{2}) \sin θ = μ (\sin \frac{π}{2} \cos r_{2} - \sin r_{2} \cos \frac{π}{2}) \sin θ = μ \cos r_{2} \dots (2)$

From above two equations,

$\sin θ = μ \sqrt{1 - \sin r_{2}^{2}} \sin θ = μ \sqrt{1 - \frac{1}{μ^{2}}} \sin θ = \sqrt{μ^{2} - 1} \Rightarrow θ = \sin^{- 1} (\sqrt{μ^{2} - 1}) \dots (3)$

Here, $μ = 1.3$ .

So, from equation $(3)$ ,

$θ = \sin^{- 1} (\sqrt{{(1.3)}^{2} - 1})$

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