Light of wavelength 600 nm (in air) falls on a slit of width 0.3 mm. The screen is 1.5 m from the slit. The whole set-up is placed in water of refractive index \(4/3\). The width of the central maximum on the screen is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4.5 mm
In water, light travels slower. Its frequency stays the same, so its wavelength becomes \( \lambda/n \). The width of the central maximum is \( 2\lambda D/a \). So the width also becomes smaller by the factor \( n \).
\( \lambda=600\ \mathrm{nm}=6\times10^{-7}\ \mathrm{m} \)
\( a=0.3\ \mathrm{mm}=3\times10^{-4}\ \mathrm{m} \)
\( D=1.5\ \mathrm{m} \)
\( n=4/3 \)
\( \begin{aligned} \lambda' &= \frac{\lambda}{n} \\ &= \frac{600\times3}{4}\ \mathrm{nm} \\ &= 450\ \mathrm{nm} \end{aligned} \)
\( \begin{aligned} w &= \frac{2\lambda' D}{a} \\ &= \frac{2\times4.5\times10^{-7}\times1.5}{3\times10^{-4}} \\ &= 4.5\times10^{-3}\ \mathrm{m}=4.5\ \mathrm{mm} \end{aligned} \)
Hence, the central maximum in water is 4.5 mm wide.
NCERT (2026-27):
The single slit – “It is observed that the intensity has a central maximum at θ = 0 and other secondary maxima at”
\( \displaystyle \theta \approx \left(n+\frac{1}{2}\right)\frac{\lambda}{a} \)
“which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a},\quad n=\pm 1,\pm 2,\pm 3,\ldots \)
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.”