Answer to Question #202077 in Optics for Prachi

Question

State and explain Fermat’s principle. Using this principle, derive Snell’s law of 

refraction.

Answer 1

Fermat’s principle states that “light travels between two points along the path that requires the least time, as compared to other nearby paths.” From Fermat’s principle, one can derive (a) the law of reflection [the angle of incidence is equal to the angle of reflection] and (b) the law of refraction [Snell’s law].

(a) Consider the light ray shown in the figure. A ray of light starting at point A reflects off the surface at point P before arriving at point B, a horizontal distance l from point A. We calculate the length of each path and divide the length by the speed of light to determine the time required for the light to travel between the two points.

$t = \frac{\sqrt{x^2+h_1^2}}{c} + \frac{\sqrt{(1-x)^2+h^2_2}}{c}$

To minimize the time we set the derivative of the time with respect to x equal to zero. We also use the definition of the sine as opposite side over hypotenuse to relate the lengths to the angles of incidence and reflection.

$0= \frac{dt}{dx} = \frac{x}{c \sqrt{x^2+h_1^2}} + \frac{-(1-x)}{c \sqrt{(1-x)^2+h^2_2}} → \\ \frac{x}{c \sqrt{x^2+h_1^2}} = \frac{(1-x)}{c \sqrt{(1-x)^2+h^2_2}} \; → \\ sinθ_1 = sinθ_2 → θ_1= θ_2$

Thus, law of reflection is proved.

$t = \frac{\sqrt{x^2+h_1^2}}{c/n_1} + \frac{\sqrt{(1-x)^2+h^2_2}}{c/n_2}$

To minimize the time we set the derivative of the time with respect to x equal to zero. We also use the definition of the sine as opposite side over hypotenuse to relate the lengths to the angles of incidence and reflection.

$0= \frac{dt}{dx} = \frac{n_1x}{c \sqrt{x^2+h_1^2}} + \frac{-n_2(1-x)}{c \sqrt{(1-x)^2+h^2_2}} → \\ \frac{n_1x}{c \sqrt{x^2+h_1^2}} = \frac{n_2(1-x)}{c \sqrt{(1-x)^2+h^2_2}} \; → \\ n_1sinθ_1 = n_2sinθ_2$

Thus, Snell law is proved