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Chapter 24: Problem 46

The separation between successive fringes in a double slit arrangement is \(x\). If the whole arrangement is dipped under water, what will be the new fringe separation? [The wavelength of light being used is \(5000 \AA\) A] (a) \(1.5 x\) (b) \(x\) (c) \(0.75 x\) (d) \(2 x\)

Short Answer

Expert verified
The new fringe separation is 0.75x, which corresponds to option (c).

Step by step solution

01

Understand the Problem

In this problem, we have a double slit arrangement initially in air, and we want to find the new fringe separation when the arrangement is submerged in water. We need to take into account the change in medium, which affects the wavelength of light.
02

Recall the Formula for Fringe Separation

The fringe separation in a double slit experiment is given by \( x = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength of light, \( D \) is the distance to the screen, and \( d \) is the separation between slits.
03

Consider the Effect of Medium Change

When the arrangement is submerged in water, the wavelength of light changes. The wavelength of light in a medium is given by \( \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n} \), where \( n \) is the index of refraction of the medium (water in this case, \( n \approx 1.33 \)).
04

Calculate New Wavelength

The new wavelength of light in water becomes \( \lambda_{\text{water}} = \frac{5000 \ \mathring{A}}{1.33} \approx 3750 \ \mathring{A} \).
05

Determine the New Fringe Separation

Using the fringe separation formula with the new wavelength, the fringe separation in water is \( x_{\text{water}} = \frac{\lambda_{\text{water}} D}{d} = \frac{3750}{5000}x = 0.75x \).
06

Select the Correct Answer

From the options given (a, b, c, d), option (c) corresponds to our calculated fringe separation, 0.75x.

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Key Concepts

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

Fringe Separation
In the fascinating Young's Double Slit Experiment, fringe separation is a key observation that helps us understand light's wave nature. Fringe separation refers to the distance between consecutive bright or dark bands (fringes) on a screen. These are created due to the interference of light waves passing through two slits. This separation is calculated using the formula:
  • \( x = \frac{\lambda D}{d} \),
where:
  • \( \lambda \) is the wavelength of light used,
  • \( D \) is the distance from the slits to the screen,
  • \( d \) is the distance between the slits.
This relationship shows that fringe separation is directly proportional to the wavelength and the screen distance but inversely proportional to the slit separation. By understanding this, we can predict how changes in any of these variables affect the pattern we observe.
Wavelength in Different Mediums
Light behaves differently when it travels from one medium to another. This behavior is crucial in understanding the wavelength changes in Young's Double Slit Experiment when shifting mediums, such as moving from air to water. The wavelength of light in a medium is determined by the medium's index of refraction, following the equation:
  • \( \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n} \).
Here, \( \lambda_{\text{air}} \) is the wavelength of light in air and \( n \) is the medium's index of refraction.
When light enters a denser medium like water, its speed decreases, which reduces the wavelength. For example, if light with a wavelength of \( 5000 \ \AA \) in air enters water (with \( n \approx 1.33 \)), the new wavelength becomes approximately \( 3750 \ \AA \). This change in wavelength causes the fringe separation to reduce in water.
Index of Refraction
The index of refraction is a measure of how much light slows down in a given medium compared to its speed in vacuum. This concept is pivotal in optics and impacts how we perceive the changes in light's behavior in different environments. The formula for the index of refraction is:
  • \( n = \frac{c}{v} \),
where:
  • \( c \) is the speed of light in vacuum,
  • \( v \) is the speed of light in the medium.
Water, for instance, has an index of refraction of approximately 1.33, meaning light travels slower in water than in air. Understanding this property aids in explaining why light's wavelength decreases when entering water from air, impacting the separation of interference fringes in experiments like Young's Double Slit.

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Most popular questions from this chapter

Light of wavelength \(2 \times 10^{-3} \mathrm{~m}\) falls on a slit of width \(4 \times 10^{-3} \mathrm{~m}\). The angular dispersion of the central maximum will be (a) \(30^{\circ}\) (b) \(60^{\circ}\) (c) \(90^{\circ}\) (d) \(180^{*}\)

In Young's two slit experiment the distance between the two coherent sources is \(2 \mathrm{~mm}\) and the screen is at a distance of \(1 \mathrm{~m}\). If the fringe width is found to be \(0.03 \mathrm{~cm}\), then the wavelength of the light used is (a) \(4000 \mathrm{~A}\) (b) \(5000 \mathrm{~A}\) (c) \(5890 \mathrm{~A}\) (d) \(6000 \dot{A}\)

Monochromatic light of wavelength \(589 \mathrm{~nm}\) is incident from air on a water surface. What are the wavelength and speed of refracted light? Refractive index of water is \(1.33 .\) (NCERT] (a) \(4.20 \times 10^{-6}\) and \(4.0 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (b) \(3.68 \times 10^{-9} \mathrm{~m}\) and \(3.02 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (c) \(1.9 \times 10^{-10} \mathrm{~m}\) and \(3.2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) (d) \(4.42 \times 10^{-7} \mathrm{~m}\) and \(2.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

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

The fringe width at a distance of \(50 \mathrm{~cm}\) from the slits in Young's experiment for light of wavelength 6000 ? is \(0.048 \mathrm{~cm}\). The fringe width at the same distance for \(\lambda=5000 \AA\), will be (a) \(0.04 \mathrm{~cm}\) (b) \(0.4 \mathrm{~cm}\) (c) \(0.14 \mathrm{~cm}\) (d) \(0.45 \mathrm{~cm}\)
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