Question

A parallel beam of light of wavelength 4000Å passes through a slit of width 5×10^-3m. What will be the angular spread of the central maxima in the diffraction pattern ?
A. 1.6×10^-3 rad
B. 1.6×10^-4 rad
C. 1.2×10^-3 rad
D. 3.2×10^-3 rad
E. 3.2×10^-4 rad

Answer 1

Hint: As it is well known that central maxima is the brightest central portion of diffraction pattern and it has maximum intensity.It is between the 1st order minima from the centre of the screen on both sides of centre.

Formula used:
Here, formula will be:
nλ = asinθ , where n = nth fringe
λ = wavelength
a = slit width
θ = angular position of any point on the screen, taken from the slit centre
And tanθ = y/D where y = Position of minima
                                        D = Separation between slit and source

Complete Step by Step Answer:
Firstly, we are required to understand that the angle subtended by the beam on the screen will be too small that we can take θ instead of sinθ and θ instead of tanθ.
So,nλ = aθ
Here, n = 1. Therefore, θ = /a
Also, θ = y/D
So, /a = y/D
y = λD/a
Hence, Width of central maximum = 2λD/a
Angular width of central maximum = 2θ = 2λ/a
Here, we are asked to find out angular width.
a = 5×10-3m, λ = 4000Å
So, $2 \times \dfrac{4000\times10^{-10}}{5\times10^{-3}}$ = 1600×10^-7 rad = 1.6×10^-4 rad
Hence, it is proved that option (B) is the correct option.

Note: It should remain in memory that it will be twice the λ/a because this width is on both sides of the screen due to which it becomes twice of θ. Unit conversion should be in standard form.