Question

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

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

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 \[K = I_1 + I_2 + 2\sqrt{I_1 I_2}\]

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 \(\cos(2\pi N)\).

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.

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.

Key concepts

Intensity in Interference Patterns

In Young's double slit experiment, the intensity of light on a screen forms an interference pattern. This pattern consists of alternating bright and dark bands, known as fringes. The brightness or intensity at any point can be determined using the formula: \[I = I_1 + I_2 + 2\sqrt{ I_1 I_2}\cos\phi\] where \(I_1\) and \(I_2\) are the intensities from each of the slits, and \(\phi\) is the phase difference between the two waves.

• In regions where waves meet in phase, we see bright fringes due to constructive interference.
• In areas where they are out of phase, dark fringes occur, showing destructive interference.

This concept helps us understand how varying the path causes different intensities on the screen.

Path Difference Effect on Intensity

The path difference refers to the difference in distance traveled by two light waves from the slits to a point on the screen. It directly affects the interference pattern's intensity. When the path difference is a whole number multiple of the wavelength (\(\lambda\)), the waves reinforce each other, maximizing the intensity. Conversely, when the path difference is a half-wavelength multiple, the waves cancel each other, reducing the intensity to zero. This cyclical reinforcement and cancellation form the characteristic pattern observed. Understanding path differences is essential to predict where bright and dark fringes will appear.

Phase Difference Calculation

The phase difference \(\phi\) between waves is crucial for calculating interference intensity. It is expressed by: \[\phi = \frac{2\pi}{\lambda} \Delta x\] where \(\Delta x\) is the path difference. For example, if \(\Delta x = \lambda\), then \(\phi = 2\pi\). This implies a full cycle or complete constructive interference. Knowing \(\phi\) helps determine the interference type (constructive or destructive) at any given point. A complete cycle (\(2\pi\)) results in maximum brightness, while a half-cycle (\(\pi\)) leads to darkness, as the waves destructively interfere.

Interference and Superposition of Waves

Understanding the interference and superposition principle is fundamental in wave behavior study. When two waves overlap, they superimpose to form a single resultant wave. This superposition can be either constructive or destructive:

• Constructive interference: Occurs when waves align in phase, adding together to form a stronger wave, leading to bright fringes.
• Destructive interference: Happens when waves are out of phase, negating each other, resulting in dark fringes.

These concepts explain the intricate patterns formed in Young’s experiment, illustrating wave interactivity fundamentals. Mastery of wave superposition can illuminate many phenomena in optics and other wave-related fields.