Fraunhofer Diffraction by Double Slit

Describe Fraunhofer diffraction due to double slit with necessary theory and discuss the intensity distribution. What are the conditions of missing orders?

Describe the features of a double slit Fraunhofer’s diffraction pattern. What is missing order?


Fraunhofer Diffraction by Double Slit

In the double-slit diffraction experiment, the two slits are illuminated by a single light beam. If the width of the slits is small enough (less than the wavelength of the light), the slits diffract the light into cylindrical waves. These two cylindrical wavefronts are superimposed, and the amplitude, and therefore the intensity, at any point in the combined wavefronts depend on both the magnitude and the phase of the two wavefronts.

Fraunhofer Diffraction by Double Slit

If consider a monochromatic light beam incident on 2 parallel slit PQ and RS having l width. After that light focused on lens and diffraction pattern find at screen QQ’. Now we consider the S1 and S2 is the middle point of slit 1 and slit 2. Hence the resultant amplitude is Asinβ/β at a point T.

Fraunhofer Diffraction by Double Slit

Fraunhofer Diffraction by Double Slit

Problem 1: In a double slit diffraction the screen is placed 165 cm away from the slit. The width of the slits is 0.08 mm and they are 0.4 mm apart. Find the wavelength of light, if the fringe width is 0.25 cm.

Solution: Given that D = 165 cm; X = 0.25 cm; d = 0.4 mm=0.04 cm

The fringe width is given by

X = Dl/d

l = Xd/D

l = 0.25 X 0.04 /165

l = 6.061 X 10-5 cm

l = 6061 A0


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