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  • 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  <img alt="\math

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  \mathrm{\frac{d}{R}}  so that the light that enters at A will emerge at B. Refractive index of glass is \mathrm{\mu = 1.5}

Option: 1

0.25


Option: 2

0.5


Option: 3

0.75


Option: 4

0.8


Answers (1)

best_answer

Consider the figure. If smallest angle of incidence \theta is greater than critical angle then all light will emerge out at \mathrm{B}.

\mathrm{\Rightarrow \quad \theta \geq \sin ^{-1}\left(\frac{1}{\mu}\right)}
\mathrm{\Rightarrow \quad \sin \theta \geq \frac{1}{\mu}}

From figure, \mathrm{\sin \theta=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{d}}}

\mathrm{\Rightarrow \frac{R}{R+d} \geq \frac{1}{\mu} \Rightarrow\left(1+\frac{d}{R}\right) \leq \mu}
\mathrm{\Rightarrow \frac{d}{R} \leq \mu-1 \Rightarrow\left(\frac{d}{R}\right)_{\max }=0.5}

Posted by

Ritika Jonwal

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