 Solution

Monochromatic light has one frequency, or a bandwidth sufficiently narrow for the intended measurement. A lamp can illuminate the experiment through a narrow spectral filter or monochromator, followed by a small source aperture at the focus of a collimating lens. The aperture supplies spatial coherence across the slits; the lens supplies approximately parallel incident wavefronts. Two independently illuminated incoherent slits would not provide a stationary fringe pattern.

In the Fraunhofer approximation, the optical path difference at observation angle $\theta$ is $s\sin \theta$. For equal single-slit intensities ${\mathcal{I}}_s$, adding the two wave amplitudes givesThe thin-slit idealization suppresses the individual diffraction envelope. If additionally $|Y|\ll D$, then $\sin \theta \simeq Y/D$ and $\mathcal{I}(Y)\simeq 4{\mathcal{I}}_s{\cos }^2(\pi sY/(\lambda D))$. The far-field condition does not by itself impose this small observation angle.

For order $m=10$ at $Y=0.05\mathrm{m}$,The paraxial estimate is $sY/(10D)=5.88\times 10^{-7}\mathrm{m}$, consistent to the precision of the data.

Let the upper slit contain the glass plate and take positive $Y$ towards that slit. The extra optical path length is $\Delta =(n_g-n_a)t=5.0\times 10^{-5}\mathrm{m}$. Its total upper-minus-lower path difference is $\Delta -s\sin \theta$. The order-zero fringe therefore satisfiesIt moves towards the covered slit. Using $Y_0\simeq D\Delta /s=0.425\mathrm{m}$ here is a poor small-angle approximation: $30^{\circ }$ is not small. The order-zero fringe is being tracked as the optical delay is introduced; because the pattern is periodic in phase, many other bright fringes still lie near the original centre. Neglecting unequal glass transmission changes visibility but not the zero-path-difference condition.