A right - angled isoceles prism is held on the surface of a liquid composed of miscible solvents A and B of refractive index `n_A = 1.50 and n_B = 1.30 ` respectively . The refractive index of prism is `n_p = 1.5 ` and that of the liquid is given by ` N_L = C_A n_A +(1 -C_A) n_B` where ` C_A ` is the percentage of solventA in the liquid

IF ` theta_C` is the critical angle at prism - liquid interface . the plot which best represents the variation of the critical angle with the percentage of solvent is

Options

A.
B.
C.
D.

Answer

A

Explanation

`n_p sin theta_c =n_L sin 90^@ `
` theta _c = sin^(-1) (n_L)/(n_P)`
` theta_c = sin^(-1) (n_L)/(n_p)`
` theta_C = sin^(-1)"" ((C_A n_A +(1- C_A ) n_B))/(1.5 )`
` to` Graph between ` theta_C` and `C_A` will be curve of `sin^(–1), `
Check for `C_A` = 0.5, to find most appropriate graph
` theta_C = sin^(-1) ((0.5 ( 1.5 ) +0.5 (1.3) )/(1.5 ))`
` theta_C = sin^(-1) ((14 )/(15 )) ~= 69^@`
` therefore ` correct option is (A)