Obtain the equation for radius of illumination (or) Snell's window.

Explanation

The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation `n_(2) sin i = n_(2) sin r` for the refraction happening at the point B on the boundary between the two media is,
`n_(1)siini_(c)=n_(2)sin90^@`...(1)
`n_(1)sini_(c)=n_(2) because sin90^@=1`
`sini_(C)=n_(2)/n_(1)` ...(2)
From the right angle triangle `Delta`ABC,
`sini_(c)=(CR)/(AB)=R/sqrt(d^(2) + R^(2))` ...(3)
Equating the above two equation (3) and equation (2),
`R/sqrt(d^(2) + R^(2))=n_(2)/n_(1)`
Squaring on both sides, `R^(2)/(R^(2) + d^(2))=(n_(2)/n_(1))^(2)`
Taking reciprocal, `(R^(2) + d^(2))/R^(2)=(n_(1)/n_(2))^(2)`
On further simplifying,
`1+ d^(2)/R^(2)=(n_(1)/n_(2))^(2),d^(2)/R^(2)=(n_(1)/n_(2))^(2)-1, d^(2)/R^(2)=n_(1)^(2)/n_(2)^(2) -1=(n_(1)^(2) - n_(2)^(2))/n_(2)^(2)`
Again taking reciprocal and rearranging
`R^(2)/d^(2)=n_(2)^(2)/(n_(1)^(2) - n_(2)^(2)),R^(2)=d^(2)=(n_(2)^(2)/(n_(1)^(2) - n_(2)^(2)))`
The radius of illumination is,
`R=dsqrt(n_(2)^(2)/(n_(1)^(2)-n_(2)^(2)))` ...(4)
If the rarer medium outside is air, then, `n_(2)` = 1, and we can take `n_(1)`= n
`R=d(1/sqrt(n^(2)-1))` (or) `R=d/sqrt(n^(2)-1)` ...(5)