Question

What happens with the Fraunhofer single slit diffraction pattern if the whole apparatus is immersed in water?

(a) The Wavelength of light increases

(b) Width of central maximum increases

(c) Width of central maximum decreases

(d) Frequency of light decreases

This question was posed to me at a job interview.

The origin of the question is Fraunhofer Diffraction in division Optics of Engineering Physics II

Answer 1

The correct answer is (c) Width of central maximum decreases

To explain: As the whole apparatus is now immersed in water, the wavelength of the light will change.

\(\lambda^{‘}=\frac{\lambda}{\mu}\)

Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease.

Width of central maxima = \(\frac{2λ}{a}\)

Therefore, as the wavelength decreases, the width of the central maxima decreases.