Question 149: Ray Optics

Question

A monochromatic light wave with wavelength $$\lambda_{1}$$ and frequency $$v_{1}$$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $$45^{\circ}$$ and $$30^{\circ}$$ respectively, then the wavelength $$\lambda_{2}$$ and frequency $$v_{2}$$ of the refracted wave are:

Accepted Answer

When a light wave moves from one medium to another, its speed and wavelength may change, but its frequency remains constant because it is determined by the source of the light. This is because frequency depends on the oscillations of the source, which do not change when entering a different medium.
Snell's law, which describes the relationship between the angles of incidence and refraction, can be written as:
$ n_1 \sin(	heta_1) = n_2 \sin(	heta_2), $
where $n_1$ and $n_2$ are the refractive indices of the two media, and $	heta_1$ and $	heta_2$ are the angles of incidence and refraction, respectively.
The refractive index of a medium is also related to the speed of light in that medium:
$ n = \frac{c}{v}, $
where $c$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium.
From the above relations, we see that as light enters a medium with a higher refractive index (and hence a lower speed), its wavelength decreases. The frequency remains the same.
Given that the angle of incidence is $45^\circ$ and the angle of refraction is $30^\circ$, the ratio of the refractive indices is:
$ \frac{n_2}{n_1} = \frac{\sin(	heta_1)}{\sin(	heta_2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \sqrt{2}. $
This indicates that the speed of light (and hence the wavelength) decreases by a factor of $\sqrt{2}$ as it enters the second medium. The frequency remains the same.
Therefore, the correct answer is
$ \lambda_2 = \frac{1}{\sqrt{2}} \lambda_1, \quad v_2 = v_1. $

Answer

$$\lambda_{2}=\lambda_{1}, v_{2}=\frac{1}{\sqrt{2}} v_{1}$$

Answer

$$\lambda_{2}=\lambda_{1}, v_{2}=\sqrt{2} v_{1}$$

Answer

$$\lambda_{2}=\sqrt{2} \lambda_{1}, v_{2}=v_{1}$$

Answer

$$\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}, v_{2}=v_{1}$$

Ray Optics

A monochromatic light wave with wavelength $\lambda_{1} and frequency v_{1} in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45^{\circ} and 30^{\circ} respectively, then the wavelength \lambda_{2} and frequency v_{2}$ of the refracted wave are:

Choose the correct answer: