Question

FIGURE EX$34.16$shows a transparent hemisphere with radius $R$and index of refraction $\text{n.}$What is the maximum distance $d$for which a light ray parallel to the axis refracts out through the curved surface?

Short answer

A light ray parallel to the axis can refract out through the curved surface for a maximum distance$d=\frac{R}{n}$

Step-by-step solution

Concepts and Principles

Total Internal Reflection occurs at the interface when the angle of incidence reaches a critical angle when a ray travels through a material with a higher index of refraction toward a material with a lower index.

${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Given Data

• Hopefully we still have a glass spheroid and with a distance $R$and an index of refraction of$n$ .
• A ray of light is going parallel to the hemisphere's axis.

Required Data

An maximal distance d whereby the beam of light deforms out from the contour is necessary.

Solution

The parameter of the hemispheric, derived by Equation, is the lowest angle in which the laser pulse deflects through the curved surface.

${\theta }_c=\sin -1\left(\frac{n_2}{n_1}\right)$

since then we really imagine a gyrus is just in wind and it is the hemisphere's index of refraction:

${\theta }_c={\sin }^{-1}\left(\frac{1}{n}\right)$

maximum distance

An cosine of zero point could be found and used the topology as seen in Figure$\text{1:}$

$\sin {\theta }_c=\frac{d}{R}$

${\theta }_c={\sin }^{-1}\left(\frac{d}{R}\right)$

Here where the angle of inclination is similar to total distance d at which light ray refracts on around a hemisphere's circular cylinder.

Equation Comparison

Deduce the equation

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

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