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Chapter 16: Problem 2313

In Young's double slit experiment, the intensity on screen at a point where path difference is \(\lambda\), is \(\mathrm{K}\), What will be intensity at the point where path difference is \((N 4)\) (A) \((\mathrm{K} / 2)\) (B) \(2 \mathrm{~K}\) (C) \(4 \mathrm{~K}\) (D) zero

Short Answer

Expert verified
The intensity at the point where the path difference is Nλ is the same as the intensity at the point where the path difference is λ, which is K. The given choices are misleading, and none of them match the correct answer.

Step by step solution

01

Calculate intensity at path difference λ

In Young's double slit experiment, the formula for intensity at a point on the screen is \[I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi,\] where \(I_1\) and \(I_2\) are the individual intensities from the two slits, and \(\phi\) is the phase difference between the two waves arriving at the point on the screen. Given the path difference is λ, and we know that \[\phi = \frac{2\pi}{\lambda} \Delta x,\] where \(\Delta x\) is the path difference. So, substituting the values, we get \[\phi = \frac{2\pi}{\lambda} \lambda = 2\pi.\] Now we can find the intensity at this point as \[I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(2\pi).\] Since the intensity is K, and \(\cos(2\pi) = 1\), we get \[K = I_1 + I_2 + 2\sqrt{I_1 I_2}\]
02

Calculate intensity at path difference Nλ

Now, let's find the intensity at a path difference of Nλ. In this case, the phase difference is given by \[\phi = \frac{2\pi}{\lambda} N\lambda = 2\pi N.\] The intensity at this point is given by \[I' = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(2\pi N).\] Now we need to find the value of \(\cos(2\pi N)\).
03

Find the value of cos(2πN)

The value of the cosine function repeats every \(2\pi\), i.e., for any integer value of N, \[\cos(2\pi N)=1\] considering \( N\) as an integer.
04

Calculate the new intensity

Now, using the value of \(\cos(2\pi N)\), find the intensity at path difference Nλ as: \[I' = I_1 + I_2 + 2\sqrt{I_1 I_2}\] Since we've previously found that \[K = I_1 + I_2 + 2\sqrt{I_1 I_2}\] Therefore, \[I' = K\] So, the answer is: (C) \(4\mathrm{~K}\) The intensity at the point where path difference is Nλ is the same as the intensity at the point where path difference is λ, which is K. The given choices are misleading, and none of them match the correct answer.

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Most popular questions from this chapter

Mono chromatic light of wavelength \(399 \mathrm{~nm}\) is incident from air on a water \((\mathrm{n}=1.33)\) Surface. The wavelength of refracted light is \(\mathrm{nm}\) (A) 300 (B) \(\overline{600}\) (C) 333 (D) 443

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To get five images of a single object one should have two plane mirrors at an angle of (A) \(36^{\circ}\) (B) \(72^{\circ}\) (C) \(80^{\circ}\) (D) \(302^{\circ}\)
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