Question 299: Waves Optics

Question

In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the maximum intensity of light at a point on the screen is ' K ' units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ ' is $\left(\cos 60^{\circ}=\sin 30^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right)$

Accepted Answer

In Young's double-slit experiment with monochromatic light of wavelength $ \lambda $, the maximum intensity of light at a given point on the screen is represented as $ K $ units. We aim to find the intensity at a point where the path difference is $ \frac{\lambda}{6} $.
The intensity $ I $ is given by:
$ I = 4I_0 \cos^2 \frac{\phi}{2} \quad 	ext{... (i)} $
The maximum intensity, $ K = 4I_0 $, occurs when the phase difference $ \phi = 0 $.
For a path difference of $ \frac{\lambda}{6} $, the phase difference is calculated as:
$ \phi = \frac{2\pi}{\lambda} 	imes 	ext{path difference} = \frac{\pi}{3} $
Substituting this into our intensity equation, we get:
$ I = K \cos^2\left(\frac{\pi}{6}ight) = K \left(\frac{3}{4}ight) $
Therefore, the intensity at this point is:
$ I = \frac{3K}{4} $

Answer

$\frac{3 \mathrm{~K}}{4}$

Answer

$\frac{\mathrm{K}}{4}$

Answer

$\frac{\mathrm{K}}{2}$

Answer

$\mathrm{K}$

Waves Optics

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