Question 308: Waves Optics

Question

In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the intensity of light at a point on the screen where path difference ' $\lambda$ ' is K units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is $\left[\cos \frac{\pi}{6}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right]$

Accepted Answer

In Young's double slit experiment, the intensity of the light at a point on the screen depends on the path difference and can be found using the relation:
$$ I = I_0 \cos^2 \left(\frac{\Delta \phi}{2}\right) $$
where $I_0$ is the maximum intensity, and $\Delta \phi$ is the phase difference related to the path difference $\Delta x$ by:
$$ \Delta \phi = \frac{2\pi}{\lambda} \Delta x $$
For the path difference $\Delta x = \lambda$, the phase difference is:
$$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi $$
The intensity at this point is given by:
$$ I = I_0 \cos^2 \left(\frac{2\pi}{2}\right) = I_0 \cdot 1 = I_0 $$
This intensity is stated as $K$ units.
Now, for the path difference $\Delta x = \frac{\lambda}{6}$, the phase difference becomes:
$$ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3} $$
The intensity at this point is:
$$ I = I_0 \cos^2 \left(\frac{\pi}{6}\right) $$
Given that $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, we have:
$$ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} $$
Since $I_0 = K$, the intensity becomes:
$$ I = \frac{3K}{4} $$
Therefore, the correct answer is:
Option B: $$\frac{3 \mathrm{~K}}{4}$$

Answer

K

Answer

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

Answer

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

Answer

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

Waves Optics

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