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A ray of light is incident on the left vertical face of a glass cube of refractive index \mu_{2},as shown in figure. The plane of incident is the plane of the page, and the cube is surrounded by liquid \left(\mu_{1}\right).What is the largest angle of incidence \theta_{1} for which total internal reflection occurs at the top surface? 

Option: 1

\quad \sin \theta_{1}=\sqrt{\left(\frac{\boldsymbol{\mu}_{2}}{\boldsymbol{\mu}_{1}}\right)^{2}-1}


Option: 2

\quad \sin \theta_{1}=\sqrt{\left(\frac{\mu_{2}}{\mu_{1}}\right)^{2}+1}


Option: 3

\quad \sin \theta_{1}=\sqrt{\left(\frac{\mu_{1}}{\mu_{2}}\right)^{2}+1}


Option: 4

\quad \sin \theta_{1}=\sqrt{\left(\frac{\mu_{1}}{\mu_{2}}\right)^{2}-1}


Answers (1)

best_answer

Consider A Point

Applying Snell's law,

\mu_{1} \sin \theta_{1}=\mu_{2} \sin \theta_{2}\quad \ldots(1)
But \, \theta_{2}=90^{\circ}-\theta_{\mathrm{c}}
\therefore \quad \cos \theta_{2}=\sin \theta_{\mathrm{c}}=\frac{\mu_{1}}{\mu_{2}}\quad \ldots(2)

Elimination of \theta_{2} between (1) and (2), we get

\therefore \quad \sin \theta_{1}=\sqrt{\left(\frac{\mu_{2}}{\mu_{1}}\right)^{2}-1}

Posted by

Pankaj

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