Question

Divice which works on rhe prinxiple of both refection and refraction

Answer 1

Reflection of light

The law of reflection of light states that the angle at which a ray is incident on a reflective surface is equal to the angle at which it is reflected from the surface. In other words, in the following diagram (light travels from A to O to B), θi=θr.

reflection

Let’s try to derive this law from Fermat’s principle. The total distance that the ray of light has to travel from A to B with one reflection at O is

L=AO¯¯¯¯¯¯¯¯+OB¯¯¯¯¯¯¯¯

=a2+x2−−−−−−√+b2+(l−x)2−−−−−−−−−−√

Now since the speed of light is constant, minimizing the time taken by light to take the path AOB is equivalent to minimizing the path itself. To do this, we set dLdx=0.

dLdx=0

⟹ddx(a2+x2−−−−−−√+b2+(l−x)2−−−−−−−−−−√)=0

⟹2x2a2+x2√+2(l−x)(−1)2b2+(l−x)2√=0

⟹xa2+x2√=(l−x)b2+(l−x)2√

⟹sinθi=sinθr

⟹θi=θr

Refraction

The law of refraction of light, or Snell’s Law states that the ratio of the sines of the angles of incidence and refraction is equal to the inverse ratio of the indices of refraction of the two media.

refraction

With reference to the image above, Snell’s Law can be mathematically stated as,

sinθisinθr=n2n1

Now to prove this using Fermat’s principle.

We know that,

index of refraction of a medium=speed of light in vacuumspeed of light in the medium

Let the speeds of light in our media be v1=cn1 and v2=cn2. Now the total time taken by light to travel from A to B with a refraction at O is

t=AO¯¯¯¯¯¯¯¯v1+OB¯¯¯¯¯¯¯¯v2

=AO¯¯¯¯¯¯¯¯×n1c+OB¯¯¯¯¯¯¯¯×n2c

=n1a2+x2√c+n2b2+(d−x)2√c

To minimize t, we’ll set dtdx=0

ddx(n1a2+x2√c+n2b2+(d−x)2√c)=0

⟹1cddx(n1a2+x2−−−−−−√+n2b2+(d−x)2−−−−−−−−−−√)=0

⟹2xn12a2+x2√+2(d−x)n2(−1)2b2+(d−x)2√=0

⟹n1xa2+x2√=n2(d−x)b2+(d−x)2√

⟹n1sinθi=n2sinθr

⟹sinθisinθr=n2n1