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Chapter 24: Problem 75

In Young's double slit experiment, the 8 th maximum with wavelength \(\lambda_{1}\) is at a distance, \(d_{1}\) from the central maximum and the 6 th maximum with wavelength \(\lambda_{2}\) is at a distance, \(d_{2}\). Then, \(d_{1} / d_{2}\) is equal to (a) \(\frac{4}{3}\left(\frac{\lambda_{2}}{\lambda_{1}}\right)\) (b) \(\frac{4}{3}\left(\frac{\lambda_{1}}{\lambda_{2}}\right)\) (c) \(\frac{3}{4}\left(\frac{\lambda_{2}}{\lambda_{1}}\right)\) (d) \(\frac{3}{4}\left(\frac{\lambda_{1}}{\lambda_{2}}\right)\)

Short Answer

Expert verified
The ratio \(d_1 / d_2\) is \(\frac{4}{3}\left(\frac{\lambda_1}{\lambda_2}\right)\) (Option b).

Step by step solution

01

Understand the Formula for Maximum Distance

In Young's double slit experiment, the position of maxima is given by the formula:\[ d = n \frac{\lambda D}{a} \]where \(d\) is the distance of the nth maximum from the central maximum, \(\lambda\) is the wavelength, \(D\) is the distance from the slits to the screen, \(a\) is the distance between the slits, and \(n\) is the order of the maximum.
02

Apply the Formula for Given Wavelengths and Orders

For \(\lambda_1\), the 8th maximum occurs at a distance \(d_1\):\[ d_1 = 8 \frac{\lambda_1 D}{a} \]For \(\lambda_2\), the 6th maximum occurs at a distance \(d_2\):\[ d_2 = 6 \frac{\lambda_2 D}{a} \]
03

Set Up the Ratio of Distances \(d_1 / d_2\)

We need to find the ratio \(\frac{d_1}{d_2}\). Substitute the expressions for \(d_1\) and \(d_2\):\[ \frac{d_1}{d_2} = \frac{8 \frac{\lambda_1 D}{a}}{6 \frac{\lambda_2 D}{a}} \]The \(D\) and \(a\) terms cancel out because they are the same for both equations, hence:\[ \frac{d_1}{d_2} = \frac{8 \lambda_1}{6 \lambda_2} \]
04

Simplify the Ratio

Simplify the ratio obtained in Step 3:\[ \frac{8 \lambda_1}{6 \lambda_2} = \frac{4 \lambda_1}{3 \lambda_2} \]This corresponds to option (b) \(\frac{4}{3}\left(\frac{\lambda_1}{\lambda_2}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interference Pattern
In Young's double slit experiment, when light passes through two closely placed slits, it forms an interference pattern on the screen behind them. This pattern results from the wave nature of light. Two coherent light waves overlap, leading to a series of bright and dark bands. The bright bands are called interference maxima, where the waves constructively interfere, while the dark bands are the interference minima, where they destructively interfere. Understanding this pattern is crucial as it demonstrates the principle of superposition and wave interference.
Here are some essential points about interference patterns:
  • The pattern is symmetrical around a central bright band called the central maximum.
  • The spacial separation of these bands depends on the light's wavelength and the experimental setup.
  • The bright bands or maxima are the points where the optical path difference is a multiple of the wavelength.
Recognizing the interference pattern allows scientists to measure small distances and precisely determine the wavelength of light in experimentation.
Optical Path Difference
Optical path difference (OPD) is a key concept in understanding interference patterns. It refers to the difference in the distance traveled by two light waves reaching a specific point after passing through the slits. This difference is crucial as it determines whether the waves will interfere constructively or destructively.
The formula for OPD is:
  • OPD = difference in path lengths traveled by the light waves from both slits to the point of observation.
  • Constructive interference occurs when OPD is an integer multiple of the wavelength ( \( n\lambda \)).
  • Destructive interference occurs when OPD is a half-integral multiple of the wavelength ( \( (n+0.5)\lambda \)).
Hence, the OPD determines the position of maxima and minima in the interference pattern. In Young鈥檚 experiment, controlling this path difference allows for the prediction and explanation of the observed positions of maxima and minima.
Wavelength and Frequency Relationship
Wavelength and frequency are fundamental properties of waves, including light waves used in Young's double slit experiment. They are inversely proportional, meaning if one increases, the other decreases. This relationship can be expressed with the formula:
  • \( c = \lambda \times f \)
  • Where \( c \) is the speed of light, constant in a vacuum (~\( 3 \times 10^8 \) meters/second).
  • \( \lambda \) (lambda) is the wavelength of the light.
  • \( f \) represents frequency.
The understanding of this relationship helps explain the behavior of the interference pattern. Different wavelengths will lead to maxima appearing at different places since wavelength directly affects the optical path difference.
In experiments like Young鈥檚, using light of different wavelengths ( \( \lambda_1 \) and \( \lambda_2 \)) leads to distinct interference patterns. Knowing how to manipulate and measure these patterns enables scientists to explore properties of light and further confirm wave-like behaviors.

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Most popular questions from this chapter

Two slits separated by a distance of \(1 \mathrm{~mm}\) are illuminated with red light of wavelength \(6.5 \times 10^{-7} \mathrm{~m}\). The interference fringes are observed on a screen place \(1 \mathrm{~m}\) from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) \(0.65 \mathrm{~mm}\) (b) \(1.63 \mathrm{~mm}\) (c) \(3.25 \mathrm{~mm}\) (d) \(4.88 \mathrm{~mm}\)

The equations of displacement of two waves are given as \(y_{1}=10 \sin (3 \pi t+\pi / 3)\) $$ y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) $$ then what is the ratio of their amplitude? (a) \(1: 2\) (b) \(2: 1\) (c) \(1: 1\) (d) None of these

Two waves of same frequency and same amplitude from two monochromatic source are allowed to superpose at a certain point. If in one case the phase difference is \(0^{\circ}\) and in other case is \(\pi / 2\), the ratio of the intensities in the two cases will be (a) \(1: 1\) (b) \(2: 1\) (c) \(4: 1\) (d) None of these

A parallel beam of light of intensity \(I_{0}\) is incident on a glass plate, \(25 \%\) of light is reflected by upper surface and \(50 \%\) of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is (a) \(\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{2}\) (b) \(\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^{2}\) (c) \(\frac{5}{8}\) (d) \(\frac{8}{5}\)

Light of wavelength \(6000 \AA\) is incident on a single slit. The first minimum of the diffraction pattern is obtained at \(4 \mathrm{~mm}\) from the centre. The screen is at a distance of \(2 \mathrm{~m}\) from the slit. The slit width will be (a) \(0.3 \mathrm{~mm}\) (b) \(0.2 \mathrm{~mm}\) (c) \(0.15 \mathrm{~mm}\) (d) \(0.1 \mathrm{~mm}\)
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