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Question: In YDSE, $\lambda$ = 500 nm, d = 1.00 mm and D = 1.0 m. Find minimum distance from the central maxim...

In YDSE, λ\lambda = 500 nm, d = 1.00 mm and D = 1.0 m. Find minimum distance from the central maxima for which the intensity is half of the maximum intensity.

A

2.5 ×10−4\times 10^{-4}m

B

1.25 ×10−4\times 10^{-4}m

C

3 ×10−6\times 10^{-6}m

D

4.3 ×10−5\times 10^{-5}m

Answer

1.25 ×10−4\times 10^{-4}m

Explanation

Solution

The intensity II at a distance yy from the central maxima in Young's Double Slit Experiment (YDSE) is given by: I=Imaxcos⁡2(ϕ2)I = I_{max} \cos^2\left(\frac{\phi}{2}\right) where ImaxI_{max} is the maximum intensity and ϕ\phi is the phase difference.

The phase difference ϕ\phi is related to the path difference Δx\Delta x by: ϕ=2πλΔx\phi = \frac{2\pi}{\lambda} \Delta x

For a point at a distance yy from the central maxima, the path difference is: Δx=ydD\Delta x = \frac{yd}{D} where dd is the distance between the slits and DD is the distance from the slits to the screen.

Substituting Δx\Delta x into the phase difference formula: ϕ=2πλydD\phi = \frac{2\pi}{\lambda} \frac{yd}{D}

The intensity formula can be rewritten in terms of yy: I(y)=Imaxcos⁡2(πydλD)I(y) = I_{max} \cos^2\left(\frac{\pi yd}{\lambda D}\right)

We are given that the intensity is half of the maximum intensity, so I=12ImaxI = \frac{1}{2} I_{max}. 12Imax=Imaxcos⁡2(πydλD)\frac{1}{2} I_{max} = I_{max} \cos^2\left(\frac{\pi yd}{\lambda D}\right) cos⁡2(πydλD)=12\cos^2\left(\frac{\pi yd}{\lambda D}\right) = \frac{1}{2}

Taking the square root of both sides: cos⁡(πydλD)=±12\cos\left(\frac{\pi yd}{\lambda D}\right) = \pm \frac{1}{\sqrt{2}}

For the minimum distance from the central maxima, we consider the smallest positive value for the argument of cosine. Thus, we take: cos⁡(πydλD)=12\cos\left(\frac{\pi yd}{\lambda D}\right) = \frac{1}{\sqrt{2}} This implies: πydλD=π4\frac{\pi yd}{\lambda D} = \frac{\pi}{4}

Solving for yy: y=λD4dy = \frac{\lambda D}{4d}

Now, we plug in the given values: λ=500 nm=500×10−9 m\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} d=1.00 mm=1.00×10−3 md = 1.00 \text{ mm} = 1.00 \times 10^{-3} \text{ m} D=1.0 mD = 1.0 \text{ m}

y=(500×10−9 m)×(1.0 m)4×(1.00×10−3 m)y = \frac{(500 \times 10^{-9} \text{ m}) \times (1.0 \text{ m})}{4 \times (1.00 \times 10^{-3} \text{ m})} y=500×10−94×10−3 my = \frac{500 \times 10^{-9}}{4 \times 10^{-3}} \text{ m} y=5004×10−9−(−3) my = \frac{500}{4} \times 10^{-9 - (-3)} \text{ m} y=125×10−6 my = 125 \times 10^{-6} \text{ m} y=1.25×10−4 my = 1.25 \times 10^{-4} \text{ m}