Question

An electromagnetic wave, whose wave normal makes an angle of \( 45^\circ \) with the vertical, is travelling in air and strikes a horizontal liquid surface. While travelling through the liquid, it gets deviated by \( 15^\circ \). If the speed of electromagnetic wave in air is \( 3\times10^8 \text{ m/s} \), then the speed of electromagnetic wave in the liquid will be

• A. \( \frac{\sqrt{2}}{3}\times10^8 \text{ m/s} \)
• B. \( 1.5\times10^8 \text{ m/s} \)
• C. \( 2.1\times10^8 \text{ m/s} \)
• D. \( 2.5\times10^8 \text{ m/s} \)

Hint:

Remember that \( \frac{1}{\sqrt{2}} \approx 0.707 \). To quickly compute \( \frac{3}{\sqrt{2}} \), you can multiply \( 3 \times 0.707 = 2.12 \). Performing such simple mental calculations helps to find the correct decimal option instantly without extensive division.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Refraction occurs when an electromagnetic wave transitions from one medium to another, resulting in a change in speed and direction.
The angle of incidence \(i\) is measured from the normal (vertical) to the surface.
Deviation \(\delta\) is the angle between the original path of light and the refracted path.
The angle of refraction \(r\) can be calculated from the deviation.
Snell's Law relates the angles to the speeds of light in the respective media.
Step 2: Key Formula or Approach:
1. Angle relations: \(\delta = i - r\) (when entering a denser medium).
2. Snell's Law in terms of speeds: \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2}\).
3. Given: \(i = 45^\circ\), \(\delta = 15^\circ\), \(v_{air} = 3 \times 10^8 \text{ m/s}\).
Step 3: Detailed Explanation:
The angle of incidence \(i = 45^\circ\).
The wave gets deviated by \(15^\circ\) towards the normal (as it enters a liquid from air).
Angle of refraction \(r = i - \delta = 45^\circ - 15^\circ = 30^\circ\).
Now, apply Snell's Law to find the speed in the liquid \(v_2\):
\[ \frac{\sin i}{\sin r} = \frac{v_{air}}{v_{liquid}} \]
\[ v_{liquid} = v_{air} \frac{\sin r}{\sin i} \]
Substituting the values:
\[ v_{liquid} = (3 \times 10^8) \frac{\sin 30^\circ}{\sin 45^\circ} \]
\[ v_{liquid} = 3 \times 10^8 \times \frac{0.5}{1/\sqrt{2}} = 3 \times 10^8 \times 0.5 \times \sqrt{2} \]
\[ v_{liquid} = 1.5 \times \sqrt{2} \times 10^8 \text{ m/s} \]
Using \(\sqrt{2} \approx 1.414\):
\[ v_{liquid} \approx 1.5 \times 1.414 \times 10^8 \approx 2.121 \times 10^8 \text{ m/s} \]
Rounding to the given options, we get \(2.1 \times 10^8 \text{ m/s}\).
Step 4: Final Answer:
By determining the angle of refraction and applying Snell's Law, the speed of the wave in the liquid is found to be approximately \(2.1 \times 10^8 \text{ m/s}\).