Question 163: Waves Optics

Question

A single slit of width b is illuminated by a coherent monochromatic light of wavelength  $$\lambda $$. If the second and fourthminima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)

Accepted Answer

For single slit diffraction,  sin$$	heta $$ = $${{n\lambda } \over b}$$
From central maxima the position of n
th
minima = $${{n\lambda D} \over b}$$
Now when,
n = 2, then x
2
= $${{2\lambda D} \over b}$$ = 0.03 . . . .(1)
n = 4, then x
4
= $${{4\lambda D} \over b}$$ = 0.06 . . . .(2)
Performing (2) $$-$$ (1) we get,
x
4
$$-$$ x
2
= $${{2\lambda D} \over b}$$ = 0.03
$$	herefore\,\,\,$$ $${{\lambda D} \over b}$$ = $${{0.03} \over 2}$$
As, width of crntral maximum
= $${{2\lambda D} \over b}$$
= 2 $$ 	imes $$ $${{0.03} \over 2}$$
= 0.03 m
= 3 cm

Answer

1.5 cm

Answer

3.0 cm

Answer

4.5 cm

Answer

6.0 cm

Waves Optics

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