Q.Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.

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 $\dfrac{I_1}{I_2} = 25$.

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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 $a \propto \sqrt I$:

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

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

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

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

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