Question

A monochromatic light wave with wavelength λ₁ and frequency ν₁, in air, enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength λ₂ and frequency ν₂ of the refracted wave are:

• A. \(\lambda _{1}=\frac{\lambda _{2}}{\sqrt{2}},v_{1}=v_{2}\)
• B. \(\lambda _{1}=\lambda _{2},v_{1}=\frac{v_{2}}{\sqrt{2}}\)
• C. \(\lambda _{1}=\lambda _{2},v_{2}=v_{1}\)
• D. \(\lambda _{2}=\frac{\lambda _{1}}{\sqrt{2}},v_{2}=v_{1}\)

Answer 1

The Correct Option is D

\(1 sin 45^{o}=\mu sin30^{o}\)
\(\mu=\sqrt{2}\)
\(\lambda _{2}=\frac{\lambda _{1}}{\sqrt{2}}\) and the frequency does not change along with the medium.

So, the correct answer is (D): \(\lambda _{2}=\frac{\lambda _{1}}{\sqrt{2}},v_{2}=v_{1}\)

Answer 2

Refraction and Wavelength Problem

Step 1: Snell's Law

Snell's law relates the angles of incidence and refraction to the refractive indices of the two media:

\( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)

Here, \( n_1 = 1 \) (air), \( \theta_1 = 45^\circ \), \( \theta_2 = 30^\circ \). Let \( n_2 = \mu \) (refractive index of the second medium).

\( 1 \times \sin 45^\circ = \mu \sin 30^\circ \)

\( \frac{1}{\sqrt{2}} = \mu \times \frac{1}{2} \)

\( \mu = \sqrt{2} \)

Step 2: Refractive Index and Wavelength

The refractive index of a medium is related to the speed of light in the medium (\( v \)) and the speed of light in vacuum (\( c \)) by:

\( \mu = \frac{c}{v} \)

Also, the speed of light in a medium is related to its wavelength (\( \lambda \)) and frequency (\( \nu \)) by \( v = \lambda \nu \). Therefore:

\( \mu = \frac{c}{v} = \frac{\lambda_1 \nu_1}{\lambda_2 \nu_2} \)

Since frequency remains constant, we have \( \nu_1 = \nu_2 \). We also have:

\( \frac{\mu_1}{\mu_2} = \frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} \)

Step 3: Calculate λ2

Since \( \mu_1 = 1 \) (air) and \( \mu_2 = \mu = \sqrt{2} \):

\( \frac{1}{\sqrt{2}} = \frac{\lambda_2}{\lambda_1} \)

\( \lambda_2 = \frac{\lambda_1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \lambda_1 \)

Step 4: Frequency

The frequency of light remains constant when it passes from one medium to another. Therefore, \( \nu_2 = \nu_1 \).

Conclusion:

\( \lambda_2 = \frac{1}{\sqrt{2}} \lambda_1 \) and \( \nu_2 = \nu_1 \) (Option 4).