Question

FIGURE $EX34.16$shows a transparent hemisphere with radius $R$and index of refraction localid="1651380892242" $n$What is the maximum distance localid="1651380447864" $d$for which a light ray parallel to the axis refracts out through the curved surface?

Short answer

The maximum distance for which a light ray parallel to the axis refracts out through the curved surface is$d=\frac{R}{n}$.

Step-by-step solution

Critical angle

The value of the critical angle is,

${\mathrm{\theta }}_{\mathrm{\epsilon }}={\sin }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)\dots ..\left(1\right)$

The refractive index of the medium that the light rays are emitted from is$n_1$.

The $n_2$is if the angle of incidence is less than, the refractive index of the other material into which light may enter ${\theta }_e$.

Diagram for ray travel

Explanation for diagram

The beam is shown at a distance $d$from the axis, forming an angle ${\theta }_c$at the hemisphere's curved surface.

Because the angle of incidence increases as you move farther, all photons at and beyond this distance will not refract through the curved surface due to total internal reflection.

As a result, $d$is the greatest distance at which parallel light rays refract out through the curved surface.

Calculation for maximum distance

For right angle triangle,

$\sin {\theta }_e=\frac{d}{R}.....2$

Comparison of equation $1$and $2$

We get,

$\frac{d}{R}=\frac{n_2}{n_1}$

role="math" localid="1651381150857" $d=\frac{n_2}{n}\times R$

Assume that the glass hemisphere is suspended in mid-air.

Substitute $1.0$ for $n_2$and $n$for $n_1$,

$d=\frac{1}{n}\times R$

$=\frac{R}{n}$