Question

In Young’s double slit experiment deduce the condition for (i) constructive and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position 'X' on the screen.

Answer 1

Let S₁, S₂ be the two fine slits illuminated by a monochromatic source S of wavelength A.

Intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between S₂P and S₁P.

Let O be a point on the screen equidistant from S₁ and S₂, so path difference S₁O and S₂O is zero. Hence this point will have maximum intensity due to constructive interference and is called central fringe.

Let point P be lying on the screen at a distance y from O as shown in Fig.

∴ Path difference between two waves reaching at P.

= S₂P - S₁P

In rt. ∠∆ S₂BP, we have

(S₂P)² = (S₂B)² + (PB)²

(i) For construction interference [Bright Fringes]. For bright fringes, the path difference should be equal to integral multiple of d.

(ii) For destructive interference [Dark Fringes]. For dark fringes, the path difference should be an odd multiple of λ/2.

\(\frac{D}{d}\)y = (2n - 1) λ/2

or y = \(\frac{D}{d} \) (2n - 1) \(\frac{\lambda}{2}\)

If d = 0, ß = ∞ i.e. dark and bright bands will be infinitely well spaced and there will be uniform illumination.

Conclusions

1. ß does not depend upon n so that all fringes have the same width.

2. For fixed wavelength λ and separation d fringe width ß increases as D increases. Therefore, the dark fringes become darker and bright becomes less brighter. If D is very large, the two types of fringes become indistinguishable giving a sense of general illumination. Therefore, for good fringes the screen should not be too far off.

3. For fixed D and λ, the fringe width ß is more if d decreases. This is why the two sources are kept very close for good fringes.

4. Separation ∆ß between two adjacent bright or dark fringes subtend angle ∆θ at the centre O of the double slit

∆θ = \(\frac{\beta}{D} = \frac{\lambda}{d}\)

The angular separation of fringes is given by \(\frac{\lambda}{d}\) and is independent of the position of the screen. It, therefore, gives a dirct way of measuring λ, the wavelength of light.

5. Interference pattern with white light. When slits are illuminated with white light, only central maxima is white (∵ at O, y = 0, n = 0 for all values of λ). All other bright lines become coloured. Coloured lines have their violet edges towards central maxima because y ∝ λ.

Note. For clear interference pattern, lines must be wide, for which d must be small. Hence the two coherent sources must be very close.