State two conditions to obtain sustained interference of light.
In Young's double-slit experiment, using light of wavelength 400 nm, interference fringes of width 'X' are obtained. The wavelength of light is increased to 600 nm and the separation between the slits is halved. If one wants the observed fringe width on the screen to be the same in the two cases, find the ratio of the distance between the screen and the plane of the interfering sources in the two arrangements.

Solution

Condition for sustained interference are:

(i) The two sources of light should emit light continuously.
(ii) The two waves must be in same phase or bear a constant phase difference.

Wavelength of light, $λ$ = 400 nm.
Width of interference fringes = x

Fringe width in first case is,
$β = \frac{Dλ}{d}$

i.e.,    $X = \frac{D \times 400}{d}$

In the second case,
Wavelength of light = 600 nm
Distance between the slits = d/2

Since, fringe width in second case is given to be the same, we have

$β = \frac{D' \times 600}{d / 2} = \frac{D' \times 1200}{d}$

Therefore, the ratio of the distance between the screen and the plane of interfering sources n the two arrangements is given as,

$\frac{D \times 400}{d} = \frac{D' \times 1200}{d}$

$\Rightarrow$    $\frac{D}{D'} = \frac{3}{1}$

i.e.,    $D : D' = 3 : 2 .$

Some More Questions From Wave Optics Chapter

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(iii) motion of source and/or observer.
(iv) wave length.
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