Q142 / 180 · COMEDK 2024 (Morning Shift) · Physics
A ray of light travelling through a medium of refractive index $\dfrac{5}{4}$ is incident on a glass of refractive index $\dfrac{3}{2}$. Find the angle of refraction in the glass, if the angle of incidence at the given medium - glass interface is $30^{\circ}$.
- A$\sin ^{-1}\left(\dfrac{1}{3}\right)$
- Correct answerB$\sin ^{-1}\left(\dfrac{5}{12}\right)$
- C$\sin ^{-1}\left(\dfrac{6}{5}\right)$
- D$\sin ^{-1}\left(\dfrac{1}{2}\right)$
Solution
Given the refractive index of the first medium $n_1 = \dfrac{5}{4}$ and the refractive index of the second medium (glass) $n_2 = \dfrac{3}{2}$.
The angle of incidence $i = 30^{\circ}$.
Using Snell's Law, $n_1 \sin(i) = n_2 \sin(r)$, where $r$ is the angle of refraction.
Substituting the given values:
$\dfrac{5}{4} \sin(30^{\circ}) = \dfrac{3}{2} \sin(r)$
Since $\sin(30^{\circ}) = \dfrac{1}{2}$, we have:
$\dfrac{5}{4} \times \dfrac{1}{2} = \dfrac{3}{2} \sin(r)$
$\dfrac{5}{8} = \dfrac{3}{2} \sin(r)$
$\sin(r) = \dfrac{5}{8} \times \dfrac{2}{3} = \dfrac{10}{24} = \dfrac{5}{12}$
Therefore, $r = \sin^{-1}\left(\dfrac{5}{12}\right)$.
Answer: $\sin ^{-1}\left(\dfrac{5}{12}\right)$
The angle of incidence $i = 30^{\circ}$.
Using Snell's Law, $n_1 \sin(i) = n_2 \sin(r)$, where $r$ is the angle of refraction.
Substituting the given values:
$\dfrac{5}{4} \sin(30^{\circ}) = \dfrac{3}{2} \sin(r)$
Since $\sin(30^{\circ}) = \dfrac{1}{2}$, we have:
$\dfrac{5}{4} \times \dfrac{1}{2} = \dfrac{3}{2} \sin(r)$
$\dfrac{5}{8} = \dfrac{3}{2} \sin(r)$
$\sin(r) = \dfrac{5}{8} \times \dfrac{2}{3} = \dfrac{10}{24} = \dfrac{5}{12}$
Therefore, $r = \sin^{-1}\left(\dfrac{5}{12}\right)$.
Answer: $\sin ^{-1}\left(\dfrac{5}{12}\right)$
Hence, the correct answer is option: 'B'.