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Chapter 23: Problem 77

\(n\) th bright fringe if red light \(\left(\lambda_{1}=7500 \AA\right.\) ) coincides with \((n+1)\) th bright fringe of green light \(\left(\lambda_{2}=6000 \AA\right.\).) The value of \(n=\) ? (a) 4 (b) 5 (c) 3 (d) 2

Short Answer

Expert verified
The value of \( n \) is 4, so the answer is (a).

Step by step solution

01

Understanding the Problem

We need to find the value of \( n \) such that the \( n \)th bright fringe from the red light (wavelength \( \lambda_1 = 7500 \text{ 脜} \)) coincides with the \( (n+1) \)th bright fringe of the green light (wavelength \( \lambda_2 = 6000 \text{ 脜} \)).
02

Formula for Bright Fringes

The formula for bright fringes in an interference pattern is given by \( d \sin \theta = n\lambda \), where \( d \) is the slit separation, \( \lambda \) is the wavelength and \( n \) is the order of the fringe. For red light, this becomes \( d \sin \theta = n \lambda_1 \) and for green light it becomes \( d \sin \theta = (n+1) \lambda_2 \).
03

Setting the Equations Equal

Since the \( n \)th bright fringe of the red light coincides with the \( (n+1) \)th bright fringe of the green light, we set the equations equal: \( n \lambda_1 = (n+1) \lambda_2 \). This can be solved for \( n \).
04

Substituting Wavelengths

Substitute the given wavelengths: \( n(7500) = (n+1)(6000) \).
05

Simplifying the Equation

Simplify the equation: \( 7500n = 6000n + 6000 \).
06

Solving for n

Solve the simplified equation: \( 1500n = 6000 \), so \( n = \frac{6000}{1500} = 4 \).

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Key Concepts

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

Bright Fringes
Bright fringes are a core concept in wave interference patterns. When light passes through a double slit, it creates a series of bright and dark regions on a screen, known as interference fringes. These bright fringes occur because the light waves from different slits meet in phase and constructively interfere with each other.

The formula used to determine the position of bright fringes is \( d \sin \theta = n\lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle regarding the central maximum, \( n \) is the order of the bright fringe, and \( \lambda \) represents the light's wavelength. The order of the fringe \( n \) indicates how many wavelengths fit into the path difference between the waves traveling from each slit to the screen.

Understanding bright fringes helps in predicting the position of light bands on a screen and is a pivotal part of studies regarding light wave behavior. The placement and pattern depend heavily on the wavelength used and the setup geometry.
Wavelength Coincidence
Wavelength coincidence in interference patterns occurs when bright fringes from waves of different wavelengths align or overlap. In our given problem, we are looking for when the \( n \)th bright fringe of red light overlaps with the \((n+1)\)th bright fringe of green light.

To find this point of coincidence, we use the condition \( n \lambda_1 = (n+1) \lambda_2 \), where \( \lambda_1 \) and \( \lambda_2 \) represent the wavelengths of red and green light, respectively. Simplifying this equation allows us to determine the specific order \( n \) where this phenomenon occurs.
  • This is especially useful in multi-wavelength applications, as it helps understand how different light components interact and align with each other.
  • Coincidental wavelengths can reveal underlying patterns or cause interference colors due to their constructive overlapping.
Order of Fringe
The order of the fringe \( n \) is crucial in determining which number of bright fringe will appear at a given position in an interference pattern. It tells us that the path difference between two interfering waves is an integer multiple of the wavelength.

In our example, the order of the fringe for red light is denoted by \( n \), while for green light, it's denoted by \( n+1 \). This means that the green light must travel one additional wavelength's worth of distance compared to the red light for their fringes to coincide.
  • The order \( n \) is useful for scientists and engineers who need to precisely calibrate instruments that use light, ensuring they measure wave interference accurately.
  • Understanding how to calculate and use \( n \) also aids in designing optical devices like spectrometers and resolving issues in optical paths.
By determining \( n \), we can harness the principles of wave interference to control and utilize light more effectively.

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Most popular questions from this chapter

In Young's double slit experiment, white light source is used to obtain a white central fringe and a few coloured fringes. Now if a filter allowing only red light is used in front of slit \(S_{1}\) and another filter allowing only blue light is used in front of second slit \(S_{2}\), then (a) only red coloured fringes will be observed (b) only blue colourd fringes will be observed (c) both red coloured and blue coloured fringes will be observed (d) interference pattern will not form on the screen

Four light sources produce the following four waves (i) \(y_{1}=a \sin \left(\omega t+\phi_{1}\right)\) (ii) \(y_{2}=a \sin 2 \omega t\) (iii) \(y_{3}=a^{\prime} \sin \left(\omega t+\phi_{2}\right)\) (iv) \(y_{4}=a^{\prime} \sin (3 \omega t+\phi)\) Superposition of which two waves give rise to interference? (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (iii) and (iv)

In Young's double slit experiment, the source \(S\) and two slits \(A\) and \(B\) are horizontal with slit \(A\) above slit \(B\). The fringes are observed on a vertical screen \(K\). The optical path length from \(S\) to \(B\) is increased very slightly (by introducing a transparent material of higher refractive index) and optical path length from \(S\) to \(A\) is not changed. As a result the fringe system on \(K\) moves (a) vertically downwards slightly (b) vertically upwards slightly (c) horizontally, slightly to the left (d) horizontally, slightly to the right

A polaroid is placed at \(45^{\circ}\) to an incoming light of intensity \(I_{0}\). Now the intensity of light passing through the polaroid after polarisation would be (a) \(I_{o}\) (b) \(I_{o} / 2\) (c) \(I_{o} / 4\) (d) zero

In Young's double slit experiment the separation between the slits is halved and the distance between the slits and screen is doubled. The fringe width is (a) unchanged (b) halved (c) doubled (d) quadrupled
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