A cubical block of glass of refractive index `n_1` is in contact with the surface of water of refractive index `n_2.` A beam of light is incident on vertical face of the block. After refraction a total internal reflection at the base and refraction at the opposite face take place. The ray emerges at angle `theta` as shown. The value of `theta` is given by

Options

A. `sin thetaltsqrt(n_1^2-n_2^2)`
B. `cos theta lt sqrt(n_1^2-n_2^2)`
C. `sin theta lt 1/sqrt(n_1^2-n_2^2)`
D. `cos theta lt 1/sqrt(n_1^2-n_2^2)`

Answer

A

Explanation

Applying Snell's law at point M, we get
`n_1=sin theta /(sin(90^@-theta_C)) or sin theta=n_1cos theta_C`
`=n_1sqrt(1-sin^2 theta_C)=n_1sqrt(1-(n_2^2)/(n_1^2))=sqrt(n_1^2-n_2^2)`

If angle of `theta` is less than this value, then angle of
incidence at N will be greater than `theta_C`. Hence , TIR
will take place at N.