Question 190: Wave Optics

Question

In a Young's double slit experiment, the intensity at a point is $$\left(\frac{1}{4}ight)^{	ext {th }}$$ of the maximum intensity, the minimum distance of the point from the central maximum is _________ $$\mu \mathrm{m}$$. (Given : $$\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$$)

Accepted Answer

In a Young's double slit experiment, the intensity at a point can be expressed as a function of the phase difference between the light arriving from the two slits. The intensity at any point on the screen is given by:
$$I = I_{	ext{max}} \cos^2 \left( \frac{\delta}{2} ight)$$
Where:
$$I_{	ext{max}}$$ is the maximum intensity.
$$\delta$$ is the phase difference between the light waves from the two slits.
Given the intensity at a point is $$\left(\frac{1}{4}ight)^{	ext {th }}$$ of the maximum intensity, we can write:
$$\frac{I}{I_{	ext{max}}} = \frac{1}{4}$$
Substituting this into the intensity equation:
$$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} ight)$$
Taking the square root of both sides, we get:
$$\cos \left( \frac{\delta}{2} ight) = \frac{1}{2}$$
The possible solutions for $$\delta$$ are:
$$\frac{\delta}{2} = \frac{\pi}{3}$$ or $$\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}ight)$$ which gives $$\delta = \frac{2\pi}{3}$$ or $$\delta = \frac{4\pi}{3}$$.
Considering the smallest phase difference, $$\delta = \frac{2\pi}{3}$$, we use the relation for the phase difference due to path difference:
$$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$$
Thus, substituting the value of $$\delta$$, we have:
$$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$$
Solving for $$\Delta x$$, we get:
$$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$$
The minimum distance of the point from the central maximum on the screen can be found using the interference equation:
$$y = \frac{\Delta x \cdot D}{d}$$
Substituting the known values:
$$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$$
Simplifying this expression:
$$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$$
Hence, the minimum distance of the point from the central maximum is:
$$200 \, \mu \mathrm{m}$$

Wave Optics

Integer / Numeric Answer Type