A rectangulat block of refractive index `mu` is placed on a printed page lying on a horizontal surface as shown in Fig. , Find the minimum value of `mu` so that the letter L on the page is not visible from any of the vertical sides.

Answer

A, D

Explanation

For a grazing incident ray at BD for which `iapprox90degree` the angle of refraction `(90-C)` is maximum. For this C is least. Let C is greater than the critical angle.
Applying Snell's law at M
`_1^2mu=(sin90degree)/(sin(90-C))implies_1^2mu=(1)/(cosC)` ....(i)
Also `_1^2mu=(1)/(sinC)`....(ii)
When C is the critical angle.
From (i) and (ii), ``(1)/(cosC)=(1)/(sinC)impliesC=45degree`
`:'``_1^2mu=(1)/(sin45degree)=sqrt2=1.41`