Question

Let \({v_1}\) be the velocity of light in air and \({v_2}\) the velocity of light in water. According to Fermat’s Principle, a ray of light will travel from a point in the air to a point in the water by a path that minimizes the time taken. Show that

\(\frac{{sin{\theta _i}}}{{sin{\theta _r}}} = \frac{{{v_1}}}{{{v_2}}}\)

where \({\theta _i}\) (the angle of incidence) and \({\theta _r}\) (the angle of refraction) are as shown. This equation is known as Snell’s Law

Short answer

It is proved that for light traveling from air to water \(\frac{{sin{\theta _i}}}{{sin{\theta _r}}} = \frac{{{v_1}}}{{{v_2}}}\) where \({v_1}\) is the velocity of light in air, \({v_2}\) is the velocity of light in water, \({\theta _i}\) is the angle of incidence and \({\theta _r}\) is the angle of refraction

Step-by-step solution

Given data

\({v_1}\) is the velocity of light in air and \({v_2}\) is the velocity of light in water.

Minimization condition

The conditions for minima of a function is:

\({\frac{{df\left( x \right)}}{{dx}}_{x = {x_{min}}}} = 0\;\;\;\;\;.....\left( 1 \right)\)

Minimum time taken for light to travel from air to water

Let the perpendicular from point A on the surface reach the surface at point D and the perpendicular from B reach the surface at E. Let DC be called x.

The time taken for light to travel from A to C is:

\(\begin{aligned}{c}t &= \frac{{AC}}{{{v_1}}} + \frac{{CB}}{{{v_2}}}\\ &= \frac{{\sqrt {A{D^2} + {x^2}} }}{{{v_1}}} + \frac{{\sqrt {B{E^2} + {{\left( {DE - x} \right)}^2}} }}{{{v_2}}}\end{aligned}\)

From equation (1), the time is minimum for:

\(\begin{aligned}{c}{\frac{{dt}}{{dx}}_{x = {x_{\min }}}} &= 0\\\frac{{2{x_{\min }}}}{{2{v_1}\sqrt {A{D^2} + x_{\min }^2} }} - \frac{{2\left( {DE - {x_{\min }}} \right)}}{{2{v_2}\sqrt {B{E^2} + {{\left( {DE - {x_{\min }}} \right)}^2}} }} &= 0\\\frac{{{x_{\min }}}}{{{v_1}\sqrt {A{D^2} + x_{\min }^2} }} &= \frac{{\left( {DE - {x_{\min }}} \right)}}{{{v_2}\sqrt {B{E^2} + {{\left( {DE - {x_{\min }}} \right)}^2}} }}\end{aligned}\)

But from the figure:

\(\begin{aligned}{c}\frac{{{x_{\min }}}}{{\sqrt {A{D^2} + x_{\min }^2} }} &= \frac{{DC}}{{AC}}\\ &= \sin {\theta _i}\end{aligned}\)

and

\(\begin{aligned}{c}\frac{{\left( {DE - {x_{\min }}} \right)}}{{\sqrt {B{E^2} + {{\left( {DE - {x_{\min }}} \right)}^2}} }} &= \frac{{CE}}{{CB}}\\ &= \sin {\theta _r}\end{aligned}\)

Thus, the minimization condition becomes \(\frac{{\sin {\theta _i}}}{{\sin {\theta _r}}} = \frac{{{v_1}}}{{{v_2}}}\).

It is proved that for light traveling from air to water\(\frac{{sin{\theta _i}}}{{sin{\theta _r}}} = \frac{{{v_1}}}{{{v_2}}}\) where \({v_1}\) is the velocity of light in air, \({v_2}\) is the velocity of light in water, \({\theta _i}\) is the angle of incidence and \({\theta _r}\) is the angle of refraction.