Question

Show that in Young’s double slit experiment for interference of light, the widths of the bright and dark fringes are equal.

Or

Derive an expression for fringe width using Young’s double slit method for interference of light. What will happen if the distance between the two slits becomes nearly zero?

Or

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.

Or

Describe Young’s double slit experiment and obtain an expression for the fringe width.

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

or (S₂P)² = D² + \((y +\frac{d}{2})^2\) ...........(1)

In rt. ∠∆ S₁AP, we have

(S₁P)² = (S₁A)² + (PA)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.

From (4) and (5), we conclude that bright and dark fringes

have equal fringe width

ß ∝ D

∝ λ

∝ \(\frac{1}{d}\) 

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.