Question

In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maxima of the single slit pattern? (d=1mm)

Answer 1

Answer:

Width of each slit in the experiment is 0.1 mm

Explanation:

As we know that the fringe width of central maximum due to diffraction is given as

$w_d = \frac{2\lambda D}{a}$

now the fringe width due to interference pattern is given as

$w_i = \frac{\lambda D}{d}$

so we will have number of fringes in each diffraction pattern maximum is given as

$N = \frac{w_d}{w_i}$

$20 = \frac{2d}{a}$

$a = \frac{2(1 mm)}{20}$

$a = 0.1 mm$

#Learn

Topic : diffraction

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