Question

You have a triangular prism made of glass of refractive index 1.60, with angles of 30906030^{\circ}-90^{\circ}-60^{\circ}. The short side is oriented vertically. A horizontal ray hits the middle of the slanted side of the prism. Draw the path of a ray as it passes into and through the prism. Determine all angles for its trip through the prism.

Solution

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Our goal is to find all the angles described in the graph below, and determine if the light will make it out of the prism after hitting the vertical side, and if it will make it out of the prism we want to find it's angle of refraction in the air.

\line(1,0){370}

From the geometry of the graph below we can see that

θ1=90°30°=60°\theta_{1}=90\text{\textdegree}- 30 \text{\textdegree}=60\text{\textdegree}

we can use this information and Snell's law to determine θ2\color{#c34632}\theta_{2}

n1sin(θ1)=n2sin(θ2)\begin{align} n_{1}\sin{(\theta_{1})}=n_{2}\sin{(\theta_{2})} \end{align}

\bullet n1=1n_{1}=1, the refractive index of the air

.\bullet n2=1.6n_{2}=1.6, the refractive index of the glass.

\bullet θ1=60°\theta_{1}=60 \text{\textdegree}, the angle of incidence.

Substitute for these values into (1)

1×sin(60°)=1.6×sin(θ2)1\times \sin{( 60 \text{\textdegree} )}=1.6\times \sin{( \theta_{2} )}

θ2=sin1(sin(60°)1.6)\theta_{2}=\sin^{-1}{\left( \frac{ \sin{(60 \text{\textdegree} )}}{1.6} \right)}

θ2=32.76°\boxed{\theta_{2}=32.76 \text{\textdegree}}

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