Question

Derive an expression for fringe width in young's double slit experiment

Answer 1

Explanation:

If d < < L then the difference in path length r1 - r2 travelled by the two rays is approximately:

r1 - r2 $\displaystyle\approx$ dsin $\displaystyle\theta$

where $\theta$ is approximately equal to the angle that the rays make relative to a perpendicular line joining the slits to the screen.

If the rays were in phase when they passed through the slits, then the condition for constructive interference at the screen is:

dsin $\displaystyle\theta$ = m$\displaystyle\lambda$ ,m = $\displaystyle\pm$ 1, $\displaystyle\pm$ 2,...

whereas the condition for destructive interference at the screen is:

dsin $\displaystyle\theta$ = (m + $\displaystyle{1\over 2}$)$\displaystyle\lambda$ ,m = $\displaystyle\pm$ 1, $\displaystyle\pm$ 2,...

The points of constructive interference will appear as bright bands on the screen and the points of destructive interference will appear as dark bands. These dark and bright spots are called interference fringes. Note:

In the case that y , the distance from the interference fringe to the point of the screen opposite the center of the slits (see Fig.22.10) is much less than L ( y < < L ), one can use the approximate formula:

sin $\displaystyle\theta$ $\displaystyle\approx$ y/L

so that the formulas specifying the y - coordinates of the bright and dark spots, respectively are:

y Bm = $\displaystyle{\frac{m\lambda L}{d}}$ brightspots

y Dm = $\displaystyle{\frac{(m+{1\over2})\lambda L}{d}}$ darkspots

The spacing between the dark spots is

$\displaystyle\Delta$y = $\displaystyle{\frac{\lambda L}{d}}$

Answer 2

Explanation:

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