Name: Production of Polarised Light
Uploaded: 2019-07-19
Duration: 3 min 1 s
Description: How is an unpolarised light polarised? And how do we check whether the light is polarised or not? Let's take a look at this video to understand it. Watch and...

Question

Find the angular width of the Fraunhofer diffraction pattern due to single slit

Accepted Answer

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance $D$ from the slit. The maxima and minima of the pattern arise from the interference of the various Huygens wavelets arising from the different portions of the slit.

Now, imagine the single slit as being made up of two adjacent slits, each of width $\frac{a}{2}$ . Since the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will then, we get the central maximum at $O$ .
For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources $A and C (or C and B)$ , which is the condition for destructive interference.

Let line CP for the first minimum subtends an angle $θ_{1}$ at the slit,
Then $△ ABE$ is a right-angled triangle similar to $△ COP$ .This means that $∠ BAE = θ_{1}$
$BE = a \sin θ_{1}$
Bur $BE = PB - PA = (PB - PC) + (PC - PA) = \frac{λ}{2} + \frac{λ}{2} = λ$ .

Or ${asinθ}_{1} = λ$ or $\sin θ_{1} = \frac{λ}{a} or θ_{1} = \frac{λ}{a}$ (θ1 is very small and in radian) ........... ...(1)

The minima are situated on both sides of the central maxima hence the angular width of the Central maxima will be,

$2 θ_{1} = \frac{2 λ}{a}$

This is the required expression.

Find the angular width of the Fraunhofer diffraction pattern due to single slit

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