A right angled prism of refractive index n has a plane of refractive index $n_1$ so that $n_1 < n$, cemented to its diagonal face. The assembly is in air. A ray is incident on AB.

Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.

Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

Q: A right angled prism of refractive index n has a plane of refractive index $n_1$ so that $n_1 < n$, cemented to its diagonal face. The assembly is in air. A ray is incident on AB. 1. Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. 2. Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

1. $\sin \text{C}=\frac{\text{n}_1}{\text{n}_2}$ $\text{In }\triangle\text{AEF}$ $(90-\text{r}_1)+45+(9-\text{C})=180$ $\Rightarrow\text{r}_1=45-\text{C}$ $\frac{\sin\text{i}}{\sin\text{r}_1}=\text{n}\sin\text{i}=\text{n}\sin\text{r}_1$ $=\text{n}\sin(45-\text{C})$ $=\text{n}(\sin45\cos\text{C}-\cos45\sin\text{C})$ $=\frac{\text{n}}{\sqrt{2}}(\cos\text{C}-\sin\text{C})$ $=\frac{\text{n}}{\sqrt{2}}\Big(\sqrt{[1-\sin^2\text{C}}]\Big)-\sin\text{C}$ $=\frac{1}{\sqrt{2}}\Big(\text{n}^2-\text{n}^2_1\Big)-\text{n}-1$ $\text{i}=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\sqrt{\text{n}^2-\text{n}^2_1-\text{n}_2}\Big)$ 2. From $\triangle\text{ ABC}$ $\text{r}_1+90^\circ+45^\circ=180^\circ$ $\Rightarrow\text{r}_1=45^\circ\frac{\sin\text{r}}{\sin\text{r}_1}$ $\sin\text{i}=\text{n}\sin\text{r}_1$ $=1.352\sin45=0.956$ $\text{i}=\sin^-1(0.956)=72.58$

Question

A right angled prism of refractive index n has a plane of refractive index $n_1$ so that $n_1 < n$, cemented to its diagonal face. The assembly is in air. A ray is incident on AB.

Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.
Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated.

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Solution

$\sin \text{C}=\frac{\text{n}_1}{\text{n}_2}$

$\text{In }\triangle\text{AEF}$
$(90-\text{r}_1)+45+(9-\text{C})=180$
$\Rightarrow\text{r}_1=45-\text{C}$
$\frac{\sin\text{i}}{\sin\text{r}_1}=\text{n}\sin\text{i}=\text{n}\sin\text{r}_1$
$=\text{n}\sin(45-\text{C})$
$=\text{n}(\sin45\cos\text{C}-\cos45\sin\text{C})$
$=\frac{\text{n}}{\sqrt{2}}(\cos\text{C}-\sin\text{C})$
$=\frac{\text{n}}{\sqrt{2}}\Big(\sqrt{[1-\sin^2\text{C}}]\Big)-\sin\text{C}$
$=\frac{1}{\sqrt{2}}\Big(\text{n}^2-\text{n}^2_1\Big)-\text{n}-1$
$\text{i}=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\sqrt{\text{n}^2-\text{n}^2_1-\text{n}_2}\Big)$

From $\triangle\text{ ABC}$

$\text{r}_1+90^\circ+45^\circ=180^\circ$
$\Rightarrow\text{r}_1=45^\circ\frac{\sin\text{r}}{\sin\text{r}_1}$
$\sin\text{i}=\text{n}\sin\text{r}_1$
$=1.352\sin45=0.956$
$\text{i}=\sin^-1(0.956)=72.58$

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