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 an figure. Light is seen form a point A at a depth 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 its.
`n_(1)sini_(c)=n_(2)sin90^(@)`
`n_(1)sini_(c)=n_(2) " " thereforesin90^(@)=1`
`sini, = (n_(2))/(n_(1))`
From the right angle triangle `Delta ABC`. `sini_(c)=(CR)/(AB)=(R)/(sqrt(d^(2)+R^(2))) " " ...(3)`
Equating tha above two equation (3) and equation (2).
`(R)/(sqrt(d^(2)+R^(2)))=(n_(2))/(n_(1))`
Suquaring on both sides. `(R^(2))/(sqrt(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)))-1:(d^(2))/(R^(2))=(n^(2))/(n_(2)^(2))-1=(n^(2)-n^(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 illumilation is,
`R=dsqrt(n_(2)^(2)/(n_(1)^(2)-n_(2)^(2)))`
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))`