Question

State Fermat’s principle. On the basis of this principle, show that when light passes
from a medium of lower refractive index to a medium of higher refractive index, it
bends towards the normal to the interface between the two media.

Answer 1

Answer on Question #74719, Physics / Optics

State Fermat's principle. On the basis of this principle, show that when light passes from a medium of lower refractive index to a medium of higher refractive index, it bends towards the normal to the interface between the two media.

Answer:

Fermat's principle states that the optical length of the path followed by light between two fixed points, A and B, is an extremum (minimal or maximal)

For a border of two mediums:

$$AB = \sqrt{y_1^2 + x^2}, CB = \sqrt{y_2^2 + (x - d)^2}$$ $$t_1 = \frac{\sqrt{y_1^2 + x^2}}{v} = \frac{n_1 \sqrt{y_1^2 + x^2}}{c}$$ $$t_2 = \frac{n_2 \sqrt{y_2^2 + (x - d)^2}}{c}$$ $$t = t_1 + t_2 = \frac{n_1 \sqrt{y_1^2 + x^2}}{c} + \frac{n_2 \sqrt{y_2^2 + (x - d)^2}}{c}$$

Extremal (Fermat's principle): $\frac{dt}{dx} = 0 \rightarrow \frac{2n_1x}{c\sqrt{y_1^2 + x^2}} = \frac{-2n_2(x - d)}{c\sqrt{y_2^2 + (x - d)^2}} \rightarrow \frac{n_1x}{\sqrt{y_1^2 + x^2}} = \frac{n_2(d - x)}{\sqrt{y_2^2 + (x - d)^2}}$

As: $\sin \alpha = \frac{x}{\sqrt{y_1^2 + x^2}}, \sin \beta = \frac{d - x}{\sqrt{y_1^2 + x^2}} \rightarrow n_1 \sin \alpha = n_2 \sin \beta$

As $n_2 > n_1 \rightarrow \alpha > \beta$ - light bends toward the normal

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