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  • In the figure shown, for an angle of incidence , at the top surface, what is the minimum refractive

In the figure shown, for an angle of incidence 45^{\circ}, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face \mathrm{AD}?

Option: 1

\frac{\sqrt{2}+1}{2}


Option: 2

\sqrt{\frac{3}{2}}


Option: 3

\sqrt{\frac{1}{2}}


Option: 4

\sqrt{2}


Answers (1)

best_answer

At point \mathrm{P},

Refractive index, \mathrm{\mu=\frac{\sin 45^{\circ}}{\sin r} \Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}\ \ \ ............(i)}
At point \mathrm{Q} , for total internal reflection, \mathrm{sin \ i ' =\frac{1}{\mu }}
From figure, \mathrm{ i '=90^{\circ}-r }
\mathrm{ \therefore \quad \sin \left(90^{\circ}-r\right)=\frac{1}{\mu} \Rightarrow \cos r=\frac{1}{\mu}\ \ \ \ ............(ii) }
\mathrm{ \text { Now, cor } r=\sqrt{1-\sin ^2 r}=\sqrt{\left(1-\frac{1}{2 \mu^2}\right)}=\sqrt{\left(\frac{2 \mu^2-1}{2 \mu^2}\right)}\ \ \ .............(iii) }
From eqs. \mathrm{ (ii) } and \mathrm{ (iii) }, we get 
\mathrm{ \frac{1}{\mu}=\sqrt{\frac{2 \mu^2-1}{2 \mu^2}}} \\ \mathrm{\Rightarrow\ Refractive\ index,\ \mu=\sqrt{\frac{3}{2}} }
For total internal reflection at vertical face \mathrm{ AD. }

Posted by

Kuldeep Maurya

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