Question
Show that if the apex angle $\Phi$ of a prism is small, an approximate value for the angle of minimum deviation is $\delta_{\min }=(n-1) \Phi$.
Step 1
The angle of minimum deviation occurs when light passes symmetrically through the prism, meaning that the angles of incidence and refraction at both faces of the prism are equal. Show more…
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The angle of deviation through a triangular prism is defined as the angle between the incident ray and the emerging ray (angle $\delta$ ). It can be shown that when the angle of incidence $i$ is equal to the angle of refraction $r^{\prime}$ for the emerging ray, the angle of deviation is at a minimum. Show that the minimum deviation angle $\left(\delta_{\min }=D\right)$ is related to the prism angle $A$ and the index of refraction $n,$ by $$ n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A} $$ [Hint: For an isosceles triangular prism, the minimum angle of deviation occurs when the ray inside the prism is parallel to the base, as shown in the figure.]
When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle $\delta,$ Fig. $32-64 .$ Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle $\phi,$ and show that the minimum deviation angle, $\delta_{\mathrm{m}},$ is related to the prism's index of refraction $n$ by $$ n=\frac{\sin \frac{1}{2}\left(\phi+\delta_{\mathrm{m}}\right)}{\sin \phi / 2} $$ $\left[\right.$ Hint: For $\theta$ in radians, $\left.(d / d \theta)\left(\sin ^{-1} \theta\right)=1 / \sqrt{1-\theta^{2}} .\right]$
A prism that has an apex angle of $50.0^{\circ}$ is made of cubic zirconia. What is its minimum angle of deviation?
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