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A single slit of width b is illuminated by a coherent monochromatic light of wavelength λ. If the second and fourth minima 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)
Option: 1 1.5 cm
Option: 2 3.0 cm
Option: 3 4.5 cm
Option: 4 6.0 cm

Answers (1)


As the distance of nth minima from the central maximum is given as

 x_{n}=\frac{D}{d}\left[n \frac{\lambda}{2}\right]

Difference between second and fourth minimum =6−3=3

i.e x_4-x_2=3

x_4=\frac{4\lambda D}{2d} \ \ and \ \ x_2=\frac{2 \lambda D}{2d} \\ \ so \ \ x_4-x_2=3=\frac{2\lambda D}{2d} \\ \Rightarrow \frac{\lambda D}{d}=3 \ cm


\text {Now, width of central maximum }=2 \times x_{1}$ $=2 \times \frac{D}{d}\left[1* \frac{\lambda}{2}\right]$ $=\frac{D \lambda}{d}$ $=3 \mathrm{cm}


Posted by

vishal kumar

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