Question

for a prism having angle 60 and refractive index is √2 the minimum deviation

Answer 1

μ=sin((σ+A)/2) ÷ sin(A/2)    .............(1)

where→σ=angle of minimum deviation
             A=angle of prism

given that : σ= To calculate
                   μ=√2
                   A=60

Since we need to find σ,we need to rearange the terms in (1) like this:

Sin(A/2)×μ=Sin((σ+A)/2))
substituting the value of A and μ we get

Sin(60/2)×√2=sin((σ+60)/2)
(1/2)×√2=sin((σ+60)/2)                    [sin30°=1/2]
1/√2=sin((σ+60)/2) 
 sin⁻¹(1/√2)=σ/2  + 30                      [shifting "sin" to LHS]
45=σ/2+30                                       [sin⁻¹(1/√2)=45°]
15=σ/2
30=σ

Therefore the angle of minimum deviation is at 30°