Derive the expression for refractive index of the material of the prism in terms of angle of the prism and angle of minimum deviation.

Explanation

Consider a principal section of prism of reflecting angle A and refractive index n placed in a air medium . A ray of light PQ is incident on thee face AB refraction takes place and finalllu emerges out with an angle `i_(2)`

From the quadrilateral AQXR
`A + angle X = 180^(@)`
`A= 180-angleX`
From the triangle QXR
`r_(1) + r_(2)= 180-anglex` .....(2)
`A= r_(1)+r_(2)` ..........(3)
From figure
`sigma - i-(1) - r-(2) -r_(2)`
`sigma= i-(1) + i_(20 = A` ........(4)
As the angle of incidnece incerses, since deviation decreases and reaches a minmum then after it increases as angle incidence increases. ,br> If `sigma= D`
`i-(10 and i_(20 = i`
Eqn (3) and (4) becomes
`A= r+r Rightarrow A = 2r`
`Righrarrow er= (A)/(2)`
`D= i+i -A Rightarrow i= (A+d)/(2)` ,br> From Snell.s low ,br> `n= (sin i )/(sinr)`
`n = (sin(( A+ D)/(2)))/(sin((A)/(2)))`