Problem 46
Question
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) \(2 x\)
Step-by-Step Solution
Verified Answer
The new fringe separation is 0.75x, which corresponds to option (c).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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, \( n \approx 1.33 \)).
4Step 4: Calculate New Wavelength
The new wavelength of light in water becomes \( \lambda_{\text{water}} = \frac{5000 \ \mathring{A}}{1.33} \approx 3750 \ \mathring{A} \).
5Step 5: Determine the New Fringe Separation
Using the fringe separation formula with the new wavelength, the fringe separation in water is \( x_{\text{water}} = \frac{\lambda_{\text{water}} D}{d} = \frac{3750}{5000}x = 0.75x \).
6Step 6: Select the Correct Answer
From the options given (a, b, c, d), option (c) corresponds to our calculated fringe separation, 0.75x.
Key Concepts
Fringe SeparationWavelength in Different MediumsIndex of Refraction
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:
- \( x = \frac{\lambda D}{d} \),
- \( \lambda \) is the wavelength of light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.
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:
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.
- \( \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n} \).
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:
- \( n = \frac{c}{v} \),
- \( c \) is the speed of light in vacuum,
- \( v \) is the speed of light in the medium.
Other exercises in this chapter
Problem 45
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In a Young's experiment, two coherent sources are placed \(0.90 \mathrm{~mm}\) aprt and the fringes are observed one metre away. If it produces the second dark
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The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is (a) Infinite (b) 5 (c)
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