Geometrical Optics_JEE_Advanced_2021_S1 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advance 2021

Paper 1

Multiple correct answers - Select all that apply

Question

A wide slab consisting of two media of refractive indices $n_1$ and $n_2$ is placed in air as shown in the figure. A ray of light is incident from medium $\mathrm{n}_1$ to $\mathrm{n}_2$ at an angle $\theta$, where $\sin \theta$ is slightly larger than $1 / n_1$. Take refractive index of air as 1 . Which of the following statement(s) is(are) correct?

Select ALL correct options:

The light ray enters air if $\mathrm{n}_2=\mathrm{n}_1$

The light ray is finally reflected back into the medium of refractive index $\mathrm{n}_1$ if $\mathrm{n}_2<\mathrm{n}_1$

The light ray is finally reflected back into the medium of refractive index $\mathrm{n}_1$ if $\mathrm{n}_2>\mathrm{n}_1$

The light ray is reflected back into the medium of refractive index n 1 if $\mathrm{n}_2=1$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

⚠ Partially correct. Some answers are missing.

Solution

$$ \begin{aligned} & \sin \theta>\frac{1}{n_1} \text { (Given) } \\ & \text { i.e. } \sin \theta>\frac{1}{n_1} \\ & n_1 \sin \theta_1=n_2 \sin \theta_2 \\ & \sin \theta_2=\frac{n_1 \sin \theta_1}{n_2} \\ & \text { If } n_1=n_2 \text { then } \theta_2=\theta_1 \\ & n_2 \sin \theta_2=(1) \sin \theta_3 \\ & \sin \theta_3=n_2 \sin \theta_2 \\ & \sin \theta_3=n_1 \sin \theta_1 \\ & \sin \theta_1=\frac{\sin \theta_3}{n_1}>\frac{1}{n_1} \\ & \operatorname{Sin} \theta_3>1 \\ & \theta_3>90^{\circ} \end{aligned} $$ This means ray cannot enter air $$ \begin{aligned} & \sin \theta_2=\frac{\sin \theta_3}{n_2}>\frac{1}{n_2} \\ & \theta_2=90^{\circ} \end{aligned} $$ It means ray is reflected back in medium-2 for surface 1 - surface 2 interface $$ \begin{aligned} & n_2 \sin \theta_2=n_1 \sin \theta_1 \\ & \sin \theta_{2 c}=\frac{n_1}{n_2} \end{aligned} $$ $\theta_{2 \text { cile }}$; critical angle for ray to enter medium-1 $$ \begin{aligned} & \theta_2<\theta_{2 C} \\ & \sin \theta_2<\sin 2 \theta_C \\ & \frac{n_1}{n_2} \sin \theta_1<\frac{n_1}{n_2} \\ & \sin \theta_1<1 \end{aligned} $$ $\theta_1<90^{\circ}$ which is true Hence ray enters medium-1 For $\mathbf{n}_{\mathbf{2}}>\mathbf{n}_1$ $$ \begin{aligned} & \frac{n_2}{n_1} \sin \theta_2>\frac{n_2}{n_1} \\ & \sin \theta_2>\frac{1}{n_2} \end{aligned} $$ For surface 2 - air interface $$ \begin{aligned} & n_2 \sin \theta_2=\sin \theta_3 \\ & \sin \theta_2=\frac{\sin \theta_3}{n_2}>\frac{1}{n_2} \\ & \theta_2>90 \end{aligned} $$ It means ray is reflected back in medium - 2 $\mathrm{n}_2 \sin \theta_2=\mathrm{n}_1 \sin \theta_1$ $$ \begin{aligned} & \sin \theta_1=\frac{\mathrm{n}_2}{\mathrm{n}_1} \sin \theta_2 \\ & \sin \theta_{2 \mathrm{C}}=\frac{\mathrm{n}_1}{\mathrm{n}_2} \cdot \theta_{2 \mathrm{C}} \rightarrow \text { critical angle } \end{aligned} $$ For ray to enter medium - 1 $$ \begin{aligned} & \theta 2<\theta 2 c \\ & \sin \theta 2<\sin \theta 2 c \\ & \frac{n_1}{n_2} \sin \theta_1<\frac{n_1}{n_2} \\ & \sin \theta_1<1 \end{aligned} $$ $\theta 1<90^{\circ}$, which is true Hence ray enters medium - 1 Let $\mathrm{n}_2=1$ $$ \begin{aligned} & n_1 \sin \theta_1=n_2 \sin \theta_2 \\ & n_2=1 \\ & n_1 \sin \theta_1=\sin \theta_2 \\ & \sin \theta_1=\frac{\sin \theta_2}{n_1}>\frac{1}{n_2} \\ & \sin \theta_2>1 \Rightarrow \theta_2>90^{\circ} \end{aligned} $$ ray is reflected back in medium-

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JEE Advance

Physics

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