Double-Slit Fringes in Water and with a Mica Sheet
In a double-slit experiment, a student uses sodium light of wavelength 589 nm, slit separation 0.40 mm, and screen distance 1.5 m. The student measures the fringe width in air as 2.2 mm. The apparatus is then placed in a transparent tank filled with water of refractive index 1.33, and the student observes that the fringes become narrower. Next, a thin mica sheet of thickness 6.0 μm and refractive index 1.50 is placed over one of the slits while the apparatus is back in air. The central bright fringe is no longer at the original position, and several fringes have shifted. The student asks whether the change in water is due to the change in frequency, and whether the mica sheet changes the fringe width or only shifts the pattern. The teacher asks for calculations and for a clear distinction between frequency, wavelength, and optical path.
1. Calculate the fringe width in air from the given data and compare with the measured 2.2 mm. Then calculate the fringe width in water.
Read the worked answer
β_air = λD/d = (589×10⁻⁹ m)(1.5 m)/(0.40×10⁻³ m) = 2.21×10⁻³ m = 2.21 mm, in agreement with the measured 2.2 mm.
In water, the frequency stays the same, but the speed and wavelength become v = c/n and λ_w = λ/n = 589/1.33 = 443 nm.
β_w = λ_wD/d = (443×10⁻⁹)(1.5)/(0.40×10⁻³) = 1.66×10⁻³ m = 1.66 mm. Equivalently β_w = β_air/n = 2.21/1.33 = 1.66 mm.
2. For the mica sheet over one slit, calculate the additional optical path introduced and the number of fringe widths by which the pattern shifts. In which direction does the central maximum move?
Read the worked answer
The mica sheet adds an optical path (n − 1)t = (1.50 − 1.00)(6.0×10⁻⁶ m) = 3.0×10⁻⁶ m in the covered slit.
This is equivalent to (n − 1)t/λ = 3.0×10⁻⁶/(589×10⁻⁹) ≈ 5.1 fringe widths.
The central maximum moves toward the slit covered by mica. The covered slit's optical path is longer, so the point of zero path difference must have a shorter geometrical path from that slit. The fringe width itself is essentially unchanged because the extra optical path is the same at all points on the screen; the whole pattern is shifted.
3. The student says that water changes the frequency of the light and that is why the fringes narrow. Evaluate this statement. What would happen to the fringe width if the entire apparatus, including the source, were immersed in water?
Read the worked answer
Water does not change the frequency of the light; frequency is determined by the source. What changes is the speed and therefore the wavelength: λ_w = λ_air/n. Since fringe width β = λD/d, the smaller wavelength in water gives narrower fringes. The student's statement is therefore wrong in its reason, although the observed narrowing is real.
If the entire apparatus, including the source, were immersed in water, the source still emits the same frequency. The light travels through water, so its wavelength is λ_air/n and the fringe width is again reduced by the factor n. If only the space between the slits and screen were water, the same conclusion holds for that region.