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A ray incident on the droplet of water at an angle of incidence i undergoes two reflections (not total) and emerges. If the deviation suffered by the ray within the drop is minimum and the refractive index of the droplet be `mu,` then show that `cos i =sqrt((u^2-1))/8.`

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`delta_("Total")=delta_("Refraction")+2delta_("Reflection")+delta_("Refraction")`
or `delta=(i-r)+2(180^@-2r)+(i-r)`
`= 360^@+2i-6r`
`=360^@+2i-6 sin^-1(sini/mu)`
For deviation to be minimum, `(d delta)/(di)=0`
By putting first derivative of `delta`(w.r.t.i) equal to
zero, we get the desired result.

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