In the figure shown, `mu_1gtmu_2.` Find minimum value of i so that TIR never takes place at P.

Answer

`beta=theta_C.`

Explanation

Let us take `beta=theta_C. Then, alpha=90^@-beta or alpha=90^@-theta_C`
Here, `sin theta_C=mu_R/mu_D=mu_2/mu_1`
`:.theta_C=sin^-1(mu_2/mu_1)`
Applying Snell's law at point Q,
`mu_2 sini=mu_1sinalpha=mu_1 sin(90^@-theta_C)`
`=mu_1 cos theta_C=mu_1 cos sin^-1(mu_2/mu_1)`
` :. i=sin^-1[(mu_1/mu_2)cos sin^-1(mu_2/mu_1)]`
Now, we can see that starting from this value of i, we get `beta=theta_C.`
If i is increased from this value, then `alpha` will also increase`(becomesgt90^@-theta_C)` and `beta` decreases from `theta_C (becomeslttheta_C)` and no TIR takes place at P.
So far no TIR condition i has to be increased from the above value. Or this is the minimum value of i.