An electromagnetic wave of frequency \( \nu \) and wavelength \( \lambda \), travelling with speed c in air, enters a glass slab of refractive index \( \mu \). What is its frequency in the glass slab?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \nu \)
The frequency of a wave is set by its source. At the surface, the incident wave drives the atoms of the glass. The atoms oscillate at its frequency and give out waves at that same frequency. So on entering a denser medium, only the speed and the wavelength decrease. The frequency stays \( \nu \).
In the glass slab the speed is:
\( \displaystyle v=\frac{c}{\mu} \)
With the frequency unchanged, the wavelength in the glass is:
\( \begin{aligned} \lambda' &= \frac{v}{\nu} \\ &= \frac{c}{\mu\nu} \\ &= \frac{\lambda}{\mu} \end{aligned} \)
Therefore, the wave keeps the frequency \( \nu \) in the glass slab, while its speed and wavelength both fall by the factor \( \mu \).
NCERT (2026-27):
Refraction of a Plane Wave – “The above equation implies that when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency ν(= v/λ) remains the same.”
Example 10.1 – “(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.”