JEE Advanced 2021 Paper 1

Given $\sin \theta > \frac{1}{n_{1}}$ i.e. $\sin \theta_i > \frac{1}{n_{1}}$From Snell's law, $n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}$ $\Rightarrow \sin \theta_{2} = \frac{n_{1} \sin \theta_{1}}{n_{2}}$ If $n_{1}=n_{2}$ then $\theta_{2}=\theta_{1}$ Again form Snell's law, $n_{2} \sin \theta_{2} = (1) \sin \theta_{3}$ $\Rightarrow \sin \theta_{3}=n_{2} \sin \theta_{2}$ or, $\sin \theta_{3}=n_{1} \sin \theta_{1}$ $\therefore \sin \theta_{1} = \frac{\sin \theta_{3}}{n_{1}} > \frac{1}{n_{1}}$ So, $\sin \theta_{3}>1$ $\therefore \theta_{3} > 90^{\circ}$ Hence ray cannot enter air if $n_{2}=n_{1}$ For $n_{2} < n_{1}$, from Snell 's law, $\sin \theta_{1} = \frac{n_{2}}{n_{1}} \sin \theta_{2} > \frac{1}{n_{1}}$ $\therefore \sin \theta_{2} > \frac{1}{n_{2}}$ Now for surface 2 - air interface $n_{2} \sin \theta_{2} = (1) \sin \theta_{3}$ $\therefore \sin \theta_{2} = \frac{\sin \theta_{3}}{n_{2}} > \frac{1}{n_{2}}$ So, $\theta_{2} > 90^{\circ}$ Therefore ray is reflected back in medium-2. Again for surface $1-2$ interface $n_{2} \sin \theta_{2}=n_{1} \sin \theta_{1}$ $\sin \theta_{2C} = \frac{n_{1}}{n_{2}}$$\theta_{2C}$ is critical angle For ray to enter medium $-1$ of refractive index $n_{1}$ $\theta_{2} < \theta_{2C} \therefore \sin \theta_{2} < \sin 2\theta_{C}$ $\frac{n_{1}}{n_{2}} \sin \theta_{1} < \frac{n_{1}}{n_{2}} \Rightarrow \sin \theta_{1} < 1$ Here, $\theta_{1} < 90^{\circ}$ So ray is reflected back in medium of refractive index $n_{1}$ For $n_{2} > n_{1}$ $\frac{n_{2}}{n_{1}} \sin \theta_{2} > \frac{n_{2}}{n_{1}}$ For surface 2 - air interface $n_{2} \sin \theta_{2}=\sin \theta_{3}$ $\sin \theta_{2} = \frac{\sin \theta_{3}}{n_{2}} > \frac{1}{n_{2}} = 0$ so $\theta_{2} > 90^{\circ}$ Hence ray is reflected back in medium-2 $n_{2} \sin \theta_{2}=n_{1} \sin \theta_{1}$ or, $\sin \theta_{1} = \frac{n_{2}}{n_{1}} \sin \theta_{2}$ $\sin \theta_{2C} - \frac{n_{1}}{n_{2}} ; \theta_{2C}$ is critical angleFor ray to enter medium-1, $\theta_{2} < \theta_{2C}$ $\sin \theta_{2} < \sin \theta_{2C}$ $\frac{n_{1}}{n_{2}} \sin \theta_{1} < \frac{n_{1}}{n_{2}} \therefore \sin \theta_{1} < 1$ So, $\theta_{1} < 90^{\circ}$ Hence ray is finally reflected back into the medium of refractive index $n_{1}$. If $n_{2}=1$ then from Snell's law, $n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}$ or $n_{1} \sin \theta_{1}=\sin \theta_{2}$ $\therefore \sin \theta_{1} = \frac{\sin \theta_{2}}{n_{1}} > \frac{1}{n_{1}}$ or $\sin \theta_{2} > 1 \Rightarrow \theta_{2} > 90^{\circ}$ Therefore the ray of light is reflected back into the medium of refractive index $n_{1}$.