(Explanation is in the attachment)

total 3 pictures

monochromatic light is used to reflect on an opaque screen through two narrow and similar slits S1 and S2.

the screen is placed at some distance from S1 and S2.. that distance is denoted by Capital "D" and the distance between S1 and S2 is denoted by the small letter "d".

secondary sources coming from S1 and S2, the crests and troughs superpose and interfere constructively along a straight line and points where these two lines meet have high intensity and are bright.

when the crests of one wave coincide with the trough of the other, it forms dark due to destructive interference.

hope it helps you

Consider, a point P on the screen at a distance y from O' (y « 0). The two light waves from S1 and S2 reach P along paths S1P and S2P, respectively. If the path difference (Δl) between S1P and S2P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S1P and S2P is a half-integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.

From above figure,

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

S₂P2 S₂F2+ PF²

S₂P = √D² + (x+d/2)²

Similarly, S₁P = √D² + (x-d/-2)²

Path difference is given by:

S₂P - S₁P = √D² + (x + d/2)² - √ D² + (x-d/2)²

Using binomial expansion,

SP – SP = D(1+ 1/2(x/d + d/2D)² + ...) - D(1 + 1/2 (x/D -d/2D)²+...)

Ignoring higher order terms,

Ax S₂P S₁P D

For constructive interference i.e. bright fringes,

nλ = xd D

αλD d

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

BX₁₁X₁₁-1 = nλD (n-1)^D/ (b)

Imax Imin (a₁ + a₂)² 9 25 = 4/1

Solving,

Ratio of slit widths,

W1 + W2 - T4G(73)

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