Q150 / 155 · WBJEE 2026 · Physics
An electromagnetic wave, whose wave normal makes an angle of $45°$ with the vertical, is travelling in air and strikes a horizontal liquid surface. While travelling through the liquid, it gets deviated by $15°$. If the speed of electromagnetic wave in air is $3 \times 10^8$ m/s, then the speed of electromagnetic wave in the liquid will be
- A$\dfrac{\sqrt{2}}{3} \times 10^8$ m/s
- B$1.5 \times 10^8$ m/s
- Correct answerC$2.1 \times 10^8$ m/s
- D$2.5 \times 10^8$ m/s
Solution
Angle of incidence, $i = 45^\circ$
Angle of deviation, $\delta = 15^\circ$
Since the wave travels from a rarer medium (air) to a denser medium (liquid), it bends towards the normal. The angle of refraction is given by:
$r = i - \delta = 45^\circ - 15^\circ = 30^\circ$
Using Snell's law:
$\dfrac{\sin i}{\sin r} = \dfrac{c}{v}$
Substituting the given values:
$\dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{3 \times 10^8}{v}$
$\dfrac{1/\sqrt{2}}{1/2} = \dfrac{3 \times 10^8}{v}$
$\sqrt{2} = \dfrac{3 \times 10^8}{v}$
$v = \dfrac{3 \times 10^8}{\sqrt{2}} = 1.5 \times \sqrt{2} \times 10^8$
$v \approx 1.5 \times 1.414 \times 10^8 \approx 2.12 \times 10^8$ m/s
Answer: $2.1 \times 10^8$ m/s
Angle of deviation, $\delta = 15^\circ$
Since the wave travels from a rarer medium (air) to a denser medium (liquid), it bends towards the normal. The angle of refraction is given by:
$r = i - \delta = 45^\circ - 15^\circ = 30^\circ$
Using Snell's law:
$\dfrac{\sin i}{\sin r} = \dfrac{c}{v}$
Substituting the given values:
$\dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{3 \times 10^8}{v}$
$\dfrac{1/\sqrt{2}}{1/2} = \dfrac{3 \times 10^8}{v}$
$\sqrt{2} = \dfrac{3 \times 10^8}{v}$
$v = \dfrac{3 \times 10^8}{\sqrt{2}} = 1.5 \times \sqrt{2} \times 10^8$
$v \approx 1.5 \times 1.414 \times 10^8 \approx 2.12 \times 10^8$ m/s
Answer: $2.1 \times 10^8$ m/s
Hence, the correct answer is option: 'C'.