Search any question & find its solution
Question:
Answered & Verified by Expert
Let the $x-z$ plane be the boundary between two transparent media. Medium 1 in $z \geq 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z < 0$ has a refractive index of $\sqrt{3}$. A ray of light in medium 1 given by the vector $\vec{A}=6 \sqrt{3} \hat{i}+8 \sqrt{3} \hat{j}-10 \hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is
Options:
Solution:
2316 Upvotes
Verified Answer
The correct answer is:
$45^{\circ}$
$45^{\circ}$
Normal to the plane is $z$-axis
$$
\begin{aligned}
& \cos \theta_1=\frac{A_z}{A}=\frac{10}{20}=\frac{1}{2}, \theta_1=60 \\
& \mu_1 \sin \theta_1=\mu_2 \sin \theta_2 \Rightarrow \sqrt{2} \times \frac{\sqrt{3}}{2}=\sqrt{3} \sin \theta_2 \Rightarrow \theta_2=45^{\circ}
\end{aligned}
$$
$$
\begin{aligned}
& \cos \theta_1=\frac{A_z}{A}=\frac{10}{20}=\frac{1}{2}, \theta_1=60 \\
& \mu_1 \sin \theta_1=\mu_2 \sin \theta_2 \Rightarrow \sqrt{2} \times \frac{\sqrt{3}}{2}=\sqrt{3} \sin \theta_2 \Rightarrow \theta_2=45^{\circ}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.