Two media A and B of refractive indices $μ_{1} = 1.5$ and $μ_{2} = 2$ are separated by x-z plane. A ray of light travels from A to B. The incident ray and the reflected ray are represented by unit vectors $\vec{u_{1}} = a \hat{i} + b \hat{j}$ and $\vec{u_{2}} = c \hat{i} + d \hat{j}$ . Then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{a}{c} = \frac{4}{3}$

$\frac{b}{d} = \frac{3}{4}$

$\frac{a}{c} = \frac{3}{4}$

$\frac{b}{d} = \frac{4}{3}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Applying Snell’s Law at the interface separating two media, we get

$μ_{1} \sin i = μ_{2} \sin r$

From the figure, we get
$(\frac{3}{2}) (\frac{a}{\sqrt{a^{2} + b^{^{2}}}}) = 2 (\frac{c}{\sqrt{c^{2} + d^{2}}}) ... (1)$
Since $a \hat{i} + b \hat{j}$ and $c \hat{i} + d \hat{j}$ are unit vectors, so we get
$\sqrt{a^{2} + b^{2}} = \sqrt{c^{2} + d^{2}} = 1$
Substituting in equation (1), we get
$\frac{a}{c} = \frac{4}{3}$
Hence, the correct answer is B