Question

Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.

Answer 1

$\huge{\underline{\underline{\bf{\red{Answer}}}}}$

The phenomenon of diffraction of light around the sharp corners of an obstacle and spreading into the regions of geometrical shadow is called diffraction.


From the diagram, approximate path difference is given by:

BN=ABsin(θ)=asin(θ)


For BN=nλ, constructive interference occurs

Hence, anλ​=sin(θn​)

This is the nth bright fringe.


tanθn​=Dyn​​


For small θn​,

sinθn​≈tanθn​

yn​=anλD​


Width of the secondary maximum,

β=yn​−yn−1​=aλD​


Following same lines, width of secondary minimum comes out to be the same.


(b)

For first maximum,

y1​=aλD​

For 590 nm,

y1,590​=2×10−6590×10−9×1.5​

y1,590​=0.4425m


For 596 nm,

y1,596​=2×10−6596×10−9×1.5​

y1,596​=0.447m


Separation is

y1,596​−y1,590​=0.0045m



$\huge{\underline{\underline{\bf{\red{Thank\:you}}}}}$