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  • A ray incident at a point at an angle of incidence of enters a glass sphere of R.I.<img alt="n=\sqrt

A ray incident at a point at an angle of incidence of 60^{\circ}enters a glass sphere of R.I.n=\sqrt{3}  and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is

Option: 1

50^{\circ}


Option: 2

60^{\circ}


Option: 3

90^{\circ}


Option: 4

40^{\circ}


Answers (1)


Refraction at P.
\mathrm{\frac{\sin 60^{\circ}}{\sin r_{1}}=\sqrt{3}}
\mathrm{ \Rightarrow \quad \sin r_{1}=\frac{1}{2}}
\mathrm{ \Rightarrow \quad r_{1}=30^{\circ}}
\mathrm{ Since\: r_{2}=r_{1}}
\mathrm{ \therefore r_{2}=30^{\circ}}

Refraction at Q, \mathrm{ \quad \frac{\operatorname{Sin} r_{2}}{\operatorname{Sin} \mathrm{i}_{2}}=\frac{1}{\sqrt{3}}}

\mathrm{ Putting \, \mathrm{r}_{2}=30^{\circ} we\: obtain \: \mathrm{i}_{2}=60^{\circ}}

Reflection at Q
\mathrm{r_{2}^{\prime} =r_{2}=30^{\circ} }

\mathrm{\therefore \quad \alpha =180^{\circ}-\left(r_{2}{ }^{\prime}+i_{2}\right) }
             \mathrm{ =180^{\circ}-\left(30^{\circ}+60^{\circ}\right) }
            \mathrm{ =90^{\circ} }

Posted by

Ramraj Saini

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