Light of wavelength 600 nm in air falls normally on a glass plate of thickness 0.6 mm. The refractive index of the glass is 1.5. The number of wavelengths of this light that fit in the thickness of the plate is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1500
In glass the speed falls to \( c/n \) while the frequency stays the same. So the wavelength in glass is \( \lambda/n \), and more waves fit in the same thickness.
\( \begin{aligned} \lambda_g &= \frac{\lambda}{n} \\ &= \frac{600}{1.5} \\ &= 400\ \mathrm{nm} \end{aligned} \)
\( \begin{aligned} N &= \frac{t}{\lambda_g} \\ &= \frac{0.6\times10^{-3}}{400\times10^{-9}} \\ &= 1500 \end{aligned} \)
Hence, 1500 wavelengths fit in the thickness of the plate.
NCERT (2026-27):
Refraction of a plane wave – “… Further, if λ1 and λ2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to λ1 then the distance AE will be equal to λ2 … thus,”
\( \displaystyle \frac{\lambda_1}{\lambda_2}=\frac{BC}{AE}=\frac{v_1}{v_2}\quad \text{or}\quad \frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2} \)\( \qquad (10.7) \)
Refraction of a plane wave – “… Now, if c represents the speed of light in vacuum, then,”
\( \displaystyle n_1=\frac{c}{v_1} \)\( \qquad (10.4) \)
“and”
\( \displaystyle n_2=\frac{c}{v_2} \)\( \qquad (10.5) \)
“are known as the refractive indices of medium 1 and medium 2, respectively. …”