A glass rod having square cross-section is bent into the shape as shown in the figure. The radius of the inner semi-circle is R and width of the rod is d. The minimum value of $\frac{d}{R}$ so that the light that enters at A will emerge at B. Refractive index of glass is $μ = 1.5$
Answer & explanation
Correct answer: option 2
Consider the figure. If smallest angle of incidence θ is greater than critical angle then all light will emerge out at B.
$⇒θ≥\sin^{-1}(\frac{1}{μ})$
$⇒\sin θ≥\frac{1}{μ}$
From figure, $\sin θ=\frac{R}{R+d}$
$⇒\frac{R}{R+d}≥\frac{1}{μ}⇒(1+\frac{d}{R})≤μ$
$⇒\frac{d}{R}≤μ-1⇒(\frac{d}{R})_{max}=0.5$
