Describe Young's double slit experiment and show that the bright fringe and dark fringe are equally spaced.

 When light waves from two illuminated slits is incident on the screen, the path traveled by each light wave is different. This path difference leads to a phase difference in the two light waves. The path difference is different for each point on the screen and hence, intensity is different for all the points. This leads to the formation and bright and dark fringes on the screen.

Consider point P on the screen as shown in the figure.

$S_2{P}^2=S_2{F}^2+P{F}^2$

$S_{2}P=\sqrt{D^2+(x+\frac{d}{2}{)}^2}$

Similarly,

$S_{1}P=\sqrt{D^2+(x-\frac{d}{2}{)}^2}$

Path difference is given by:

$S_{2}P-S_{1}P=\sqrt{D^2+(x+\frac{d}{2}{)}^2}-\sqrt{D^2+(x-\frac{d}{2}{)}^2}$

Using binomial expansion,

$S_{2}P-S_{1}P=D(1+\frac{1}{2}(\frac{x}{D}+\frac{d}{2D}{)}^2+...)-D(1+\frac{1}{2}(\frac{x}{D}-\frac{d}{2D}{)}^2+....)$

Ignoring higher order terms,

$\Delta x=S_{2}P-S_{1}P\approx \frac{xd}{D}$

For constructive interference i.e. bright fringes,

$n\lambda =\frac{xd}{D}$

$x_n=\frac{n\lambda D}{d}$

Fringe width is equal to the distance between two consecutive maxima.

$\beta =x_n-x_{n-1}=\frac{n\lambda D}{d}-\frac{(n-1)\lambda D}{d}$

$\beta =\frac{\lambda D}{d}$

(b)

$\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2{)}^2}{(a_1-a_2{)}^2}=\frac{9}{25}$

Solving, $\frac{a_1}{a_2}=\frac{4}{1}$

Ratio of slit widths, $\frac{w_1}{w_2}=\frac{a_1^2}{a_2^2}=16$

We know that fringe width is given by,
width=λDd$𝑤𝑖𝑑𝑡ℎ={\displaystyle \frac{𝜆𝐷}{𝑑}}$
Where,
λ$𝜆$= wavelength of source used
D = distance between screen and slit
d = distance between slits
From this we can conclude that our fringe width is dependent on wavelength, distance between screen and slit and distance between slit. Colour of the fringe does not depend upon these three factors.
We know that, width of dark and bright are the same and therefore the distance between two successive dark and bright will also be the same.
The Right width distance between two successive maxima or two successive minima is the same. Therefore the fringe width of dark and bright bands will be the same.