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Obtain the equation for radius of illumination (or) Snell's window. |
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Answer» Solution :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 mediumoutside 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))` 
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