θ>sin⁻¹[μsin(A-sin⁻¹(1/μ))]
θ<sin⁻¹[μsin(A-sin⁻¹(1/μ))]
θ>cos⁻¹[μsin(A+sin⁻¹(1/μ))]
θ<cos⁻¹[μsin(A+sin⁻¹(1/μ))]
Consider refraction though AB surface
Using Snell's law,
1sinθ=μsinr.. (i)
Now for ray to transmit through AC the angle of incidence should be less the
critical angle
i.e.r'<sin⁻¹(1/μ)
From triangle APQ
A+90°-r+90°-r'=180
r'=A-r;
A-r<sin⁻¹(1/μ)
From equation (i)
θ>sin⁻¹[μsin(A-sin⁻¹(1/μ))]
Option A is correct.