JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 3)
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)
1.5 cm
3.0 cm
4.5 cm
6.0 cm
Explanation
For single slit diffraction, sin$$\theta $$ = $${{n\lambda } \over b}$$
From central maxima the position of nth minima = $${{n\lambda D} \over b}$$
Now when,
n = 2, then x2 = $${{2\lambda D} \over b}$$ = 0.03 . . . .(1)
n = 4, then x4 = $${{4\lambda D} \over b}$$ = 0.06 . . . .(2)
Performing (2) $$-$$ (1) we get,
x4 $$-$$ x2 = $${{2\lambda D} \over b}$$ = 0.03
$$\therefore\,\,\,$$ $${{\lambda D} \over b}$$ = $${{0.03} \over 2}$$
As, width of crntral maximum
= $${{2\lambda D} \over b}$$
= 2 $$ \times $$ $${{0.03} \over 2}$$
= 0.03 m
= 3 cm
From central maxima the position of nth minima = $${{n\lambda D} \over b}$$
Now when,
n = 2, then x2 = $${{2\lambda D} \over b}$$ = 0.03 . . . .(1)
n = 4, then x4 = $${{4\lambda D} \over b}$$ = 0.06 . . . .(2)
Performing (2) $$-$$ (1) we get,
x4 $$-$$ x2 = $${{2\lambda D} \over b}$$ = 0.03
$$\therefore\,\,\,$$ $${{\lambda D} \over b}$$ = $${{0.03} \over 2}$$
As, width of crntral maximum
= $${{2\lambda D} \over b}$$
= 2 $$ \times $$ $${{0.03} \over 2}$$
= 0.03 m
= 3 cm
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