Young's Double Slit with a Film and Then Under Water
In a wave-optics experiment a student uses a sodium lamp of wavelength 589 nm, slits 0.40 mm apart, and a screen 1.50 m away. The measured fringe width is 2.2 mm. She then places a thin transparent film of refractive index 1.50 over the lower slit and finds the whole pattern shifted upward by 4.4 mm, with the fringe spacing unchanged. A classmate claims the film has changed the wavelength and that the measurement is now invalid. The teacher asks the pair to repeat the experiment with the entire apparatus immersed in a transparent tank of water of refractive index 1.33, keeping the film in place, and to predict both the new fringe width and the new shift before running it.
1. Verify the measured fringe width and use the observed shift to calculate the thickness of the film.
Read the worked answer
Fringe width $\beta=\lambda D/d$:
$\beta=\dfrac{589\times10^{-9}\times1.50}{0.40\times10^{-3}}=2.21\times10^{-3}\ \text{m}\approx2.2\ \text{mm}$, which confirms the measurement.
The film adds an optical path $(n-1)t$ to the lower slit, so the zero-order condition becomes
$\dfrac{yd}{D}+(n-1)t=0$, giving a shift
$y=\dfrac{(n-1)tD}{d}$.
Therefore
$(n-1)t=\dfrac{yd}{D}=\dfrac{4.4\times10^{-3}\times4.0\times10^{-4}}{1.50}=1.17\times10^{-6}\ \text{m}$,
and with $n=1.50$,
$t=\dfrac{1.17\times10^{-6}}{0.50}=2.35\times10^{-6}\ \text{m}\approx2.4\ \mu\text{m}$.
2. Respond to the classmate's claim that the film has changed the wavelength and invalidated the experiment. What does the unchanged fringe spacing tell you?
Read the worked answer
The classmate is wrong. The film does not change the wavelength of the light in air; it adds a fixed optical path difference $(n-1)t$ between the two beams, independent of wavelength to a good approximation. A constant extra path difference simply translates the whole interference pattern, which is exactly what is observed.
The unchanged fringe spacing is the diagnostic: $\beta=\lambda D/d$ depends only on wavelength, slit separation and screen distance. Since $\beta$ is still 2.2 mm, $\lambda$ is unchanged and the geometry is unchanged. Only the position of the zero-order maximum has moved, and the direction of the shift (towards the covered slit, i.e. downward here) confirms which beam has the extra optical path. The experiment remains valid and is in fact a standard way to measure film thickness or refractive index.
3. Predict the fringe width and the shift when the whole apparatus, film included, is immersed in water of refractive index 1.33. Explain any change in the film's effect.
Read the worked answer
In water the wavelength becomes $\lambda'=\lambda/n_w=589/1.33=443\ \text{nm}$, so
$\beta'=\dfrac{\lambda'D}{d}=\dfrac{2.21}{1.33}=1.66\ \text{mm}$, about a quarter smaller.
The film's extra optical path in water is not $(n_f-1)t$ but $(n_f-n_w)t$, because the film replaces water rather than air:
$(1.50-1.33)t=0.17\times2.35\times10^{-6}=4.0\times10^{-7}\ \text{m}$.
The new shift is
$y'=\dfrac{(n_f-n_w)tD}{d}=\dfrac{4.0\times10^{-7}\times1.50}{4.0\times10^{-4}}=1.5\times10^{-3}\ \text{m}\approx1.5\ \text{mm}$.
So both the spacing and the shift shrink; the shift shrinks dramatically because the film is now only slightly denser optically than its surroundings. The qualitative direction of the shift is unchanged.