Q.Plane wavefront of light of wavelength 5500 Å is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm, find the distance between the slits.

Intensity from a slit is proportional to slit width, and amplitude is proportional to the square root of intensity, so the amplitude ratio is $\sqrt{25}=5$.

In Young's double slit experiment, the intensity of light diffracted from a slit is directly proportional to its width:

$I \propto w$

Given $\dfrac{w_1}{w_2} = \dfrac{25}{1}$, so $\dfrac{I_1}{I_2} = 25$.

Since intensity is proportional to the square of amplitude, $I \propto a^2$:

$\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{25} = 5$

Slit-width ratio 4:1 gives amplitude ratio 2:1, and Imax/Imin = (a₁+a₂)²/(a₁−a₂)² = 9:1.

The light intensity transmitted by a slit is proportional to its width, so $I_1:I_2 = w_1:w_2 = 4:1$. Since amplitude $a \propto \sqrt I$:

$\dfrac{a_1}{a_2}=\sqrt{\dfrac{I_1}{I_2}}=\sqrt{4}=2\ \Rightarrow\ a_1:a_2=2:1$

In the interference pattern:

$\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}=\dfrac{(2+1)^2}{(2-1)^2}=\dfrac{9}{1}$

A thin glass plate in one beam's path adds an extra optical path length, which shifts the whole fringe pattern without changing the fringe spacing.

In Young's double slit experiment, if a thin transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the two interfering beams, light through that slit picks up an EXTRA optical path

$\Delta = (\mu - 1)t$

compared to the other beam. This introduces a constant additional path difference at every point on the screen, which shifts the entire fringe pattern (the position of the central bright fringe moves) towards the side of the slit carrying the glass plate, by an amount

$y_0 = \dfrac{D}{d}(\mu - 1)t$

Crucially, the fringe WIDTH, $\beta = \lambda D/d$, depends only on wavelength, slit separation and screen distance — none of which change by inserting the plate — so the fringe width is unaffected; only the whole pattern shifts bodily.

I_max/I_min = [(a1+a2)/(a1−a2)]²; solve for the amplitude ratio.

$\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2 = 16 \implies \frac{a_1+a_2}{a_1-a_2} = 4$

Let $k = a_1/a_2$. Then $\dfrac{k+1}{k-1} = 4 \implies k+1 = 4k-4 \implies 5 = 3k \implies k = \dfrac{5}{3}$.

So $a_1 : a_2 = 5:3$.