In a Young's double-slit experiment, light of wavelength 600 nm in air falls on two narrow slits S1…

In a Young's double-slit experiment, light of wavelength $600\ \text{nm}$ in air falls on two narrow slits S1 and S2, a distance $1.00\ \text{mm}$ apart. The screen is $1.50\ \text{m}$ from the slits. The central bright fringe sits at O, on the line that is the same distance from both slits. A thin transparent plastic sheet, of thickness $12.0\ \mu\text{m}$ and refractive index $1.50$ , is now stuck over slit S1 only, as the figure shows. Only the sheet changes the paths of the two beams. The sheet is so thin that the sideways shift of the light inside it can be ignored. Afterwards, the whole apparatus, sheet included, is put under water of refractive index $1.33$ , with the same wavelength in air and the same distances.

Plan view of a Young's double-slit experiment, not to scale. Light of wavelength 600 nm in air passes through a source slit and falls on a barrier with two slits, S1 above and S2 below, 1.00 mm apart. A thin plastic sheet of thickness 12.0 micrometres and refractive index 1.50 is stuck over S1 only. A screen stands 1.50 m behind the barrier, and the point O on the screen lies on the line midway between the slits. Some bright and dark stripes are drawn around O.

What is the fringe width in air, before the sheet is added?
With the sheet in place in air, how far does the central bright fringe move from O?
How many fringe widths is this shift?
What is the fringe width under water, with the sheet in place?
Under water, how far does the central bright fringe move from O?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The fringe width depends on the wavelength in the medium. The shift of the pattern depends on the extra optical path that the sheet adds.

(a) Fringe width in air. $\beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9}\ \text{m} \times 1.50\ \text{m}}{1.00 \times 10^{-3}\ \text{m}} = 0.90\ \text{mm}$ .

(b) Shift in air. The sheet adds an optical path $(n_s - 1)t = 0.50 \times 12.0\ \mu\text{m} = 6.0\ \mu\text{m}$ to the beam from S1. The central fringe moves towards S1 by $\dfrac{D}{d}(n_s - 1)t = \dfrac{1.50\ \text{m}}{1.00 \times 10^{-3}\ \text{m}} \times 6.0 \times 10^{-6}\ \text{m} = 9.0\ \text{mm}$ .

(c) In fringe widths. $\dfrac{9.0\ \text{mm}}{0.90\ \text{mm}} = 10.0$ . The same number comes from $\dfrac{(n_s - 1)t}{\lambda} = \dfrac{6.0\ \mu\text{m}}{600\ \text{nm}}$ .

(d) Fringe width under water. The wavelength in water is $\dfrac{600\ \text{nm}}{1.33} = 451\ \text{nm}$ , so $\beta_w = \dfrac{\beta}{n_w} = \dfrac{0.90\ \text{mm}}{1.33} = 0.68\ \text{mm}$ .

(e) Shift under water. Now the extra optical path is only $(n_s - n_w)t = 0.17 \times 12.0\ \mu\text{m} = 2.04\ \mu\text{m}$ . The condition for equal optical paths reads $n_w \dfrac{d\,x}{D} = (n_s - n_w)t$ , so $x = \dfrac{D}{n_w d}(n_s - n_w)t = 2.30\ \text{mm}$ .

Answer: (a) $0.90\ \text{mm}$ . (b) $9.0\ \text{mm}$ . (c) $10.0$ . (d) $0.68\ \text{mm}$ . (e) $2.30\ \text{mm}$ .

Check. Solve numerically for the point of the screen where the two optical paths are equal. It lies $9.00\ \text{mm}$ from O in air and $2.30\ \text{mm}$ from O under water. Under water the shift is $\dfrac{2.30\ \text{mm}}{0.68\ \text{mm}} = 3.4$ fringe widths, which equals $\dfrac{2.04\ \mu\text{m}}{600\ \text{nm}}$ .

Watch out. The shift does not shrink by the factor $n_w$ as the fringe width does, because the sheet adds a much smaller extra path under water. Using $(n_s - 1)t$ under water would give $6.8\ \text{mm}$ , which is wrong. The shift is toward the sheet, the slit that gets the extra optical path.

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