Show that for a material with refractive index $\mu\geq\sqrt{2}$, light incident at any angle shall be guided along a length perpendicular to the incident face.
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Let the ray incident on face AB at angle i, after refraction, it travels along PQ and then interact with face AC which is perpendicular to the incident face.
Any ray is guided along AC if the angle ray makes with the face AC $(\phi)$ is greater than the critical angle. From figure
$\phi+\text{r}=90^\circ,\ \therefore\ \sin\phi=\cos\text{r}\ .....(\text{i})$
If $\phi$ is the critical angle it means, $\sin\phi\geq\frac{1}{\mu}\ .....(\text{ii})$
From (i) and (ii), $\cos\text{r}\geq\frac{1}{\mu_2}\text{ or }1-\cos^2\text{r}\leq1-\frac{1}{\mu^2}$
i.e., $\sin^3\text{r}\leq\frac{1}{\mu^2}\Rightarrow\ \sin^2\text{r}\leq1-\frac{1}{\mu^2}\ .....(\text{iii})$
Applying Snell's law on face AB,
$1.\sin\text{i}=\mu\sin\text{r}$
or $\sin^2\text{i}=\mu^2\sin^2\text{r}\Rightarrow\ \sin^2\text{r}=\frac{1}{\mu^2}\sin^2\text{i}\ .....(\text{iv})$
From (i) and (ii), $\frac{1}{\mu_2}\sin^2\text{i}\leq1-\frac{1}{\mu^2}$
of $\sin^2\text{i}\leq\mu^2-1\ .....(\text{v})$
When $\text{i}=\frac{\pi}{2}$, then we have smallest angle $\phi$.
If it is greater than the critical angle, then all other angles of incidebce shall be more than the critical angle.
Thus, $1\geq\mu^2-1\text{ or }\mu^2\geq2$
$\Rightarrow\ \mu\geq\sqrt{2}$. This is the requires result.
Any ray is guided along AC if the angle ray makes with the face AC $(\phi)$ is greater than the critical angle. From figure
$\phi+\text{r}=90^\circ,\ \therefore\ \sin\phi=\cos\text{r}\ .....(\text{i})$
If $\phi$ is the critical angle it means, $\sin\phi\geq\frac{1}{\mu}\ .....(\text{ii})$
From (i) and (ii), $\cos\text{r}\geq\frac{1}{\mu_2}\text{ or }1-\cos^2\text{r}\leq1-\frac{1}{\mu^2}$
i.e., $\sin^3\text{r}\leq\frac{1}{\mu^2}\Rightarrow\ \sin^2\text{r}\leq1-\frac{1}{\mu^2}\ .....(\text{iii})$
Applying Snell's law on face AB,
$1.\sin\text{i}=\mu\sin\text{r}$
or $\sin^2\text{i}=\mu^2\sin^2\text{r}\Rightarrow\ \sin^2\text{r}=\frac{1}{\mu^2}\sin^2\text{i}\ .....(\text{iv})$
From (i) and (ii), $\frac{1}{\mu_2}\sin^2\text{i}\leq1-\frac{1}{\mu^2}$
of $\sin^2\text{i}\leq\mu^2-1\ .....(\text{v})$
When $\text{i}=\frac{\pi}{2}$, then we have smallest angle $\phi$.
If it is greater than the critical angle, then all other angles of incidebce shall be more than the critical angle.
Thus, $1\geq\mu^2-1\text{ or }\mu^2\geq2$
$\Rightarrow\ \mu\geq\sqrt{2}$. This is the requires result.
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