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Chapter 27: Problem 4

The dark fringe for \(m=0\) in a Young's double-slit experiment is located at an angle of \(\theta=15^{\circ} .\) What is the angle that locates the dark fringe for \(m=1 ?\)

Short Answer

Expert verified
The angle for the dark fringe at m=1 is approximately 51.2掳.

Step by step solution

01

Understand the Problem

In a Young's double-slit experiment, the dark fringes occur when the path difference between the two waves is equal to a multiple of the wavelength, resulting in destructive interference. The formula for the angle of dark fringes is given by \( d \sin \theta = (m + 0.5) \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle, \( m \) is the order of the dark fringe, and \( \lambda \) is the wavelength.
02

Identify Known Values for m=0

For the first dark fringe (\(m=0\)), we know the angle is \(\theta = 15^\circ\). The formula becomes \( d \sin 15^\circ = (0 + 0.5) \lambda \). This equation allows us to relate the slit separation \(d\) and wavelength \(\lambda\).
03

Calculate Using the Formula for m=1

For \(m=1\), we need to find \(\theta\) for \( d \sin \theta = (1 + 0.5) \lambda \). Substitute the equation from Step 2 into this equation: \( d \sin \theta = 1.5 \lambda \).
04

Simplify Using Known Values

Using the proportion from Step 2, \( \sin \theta = \frac{1.5}{0.5} \sin 15^\circ \). This simplifies to \(\sin \theta = 3 \sin 15^\circ\).
05

Solve for Theta

Compute the values: \(\sin 15^\circ \approx 0.2588\), so \(\sin \theta = 3 \times 0.2588 \approx 0.7764\). Now, find \(\theta\) by calculating \(\theta = \sin^{-1}(0.7764)\).
06

Compute the Final Angle

Finally, use a calculator to find \(\theta \approx 51.2^\circ\). Therefore, the angle for the dark fringe at \(m=1\) is approximately \(51.2^\circ\).

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Key Concepts

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

Destructive Interference
In Young's double-slit experiment, destructive interference plays a key role in the formation of dark fringes. This phenomenon occurs when waves from two slits travel different paths and meet out of phase, effectively canceling each other out. For this to happen, the path difference should be an odd multiple of half the wavelength. This can be mathematically expressed as:
  • Path difference = \( (m + 0.5) \lambda \)
Where \( m \) is an integer representing the order of the dark fringe.
This results in no light intensity being observed at those points, creating the appearance of dark bands on the screen.
Understanding this condition is crucial for predicting where these dark spots will appear in the interference pattern.
Angle of Dark Fringes
The angle at which dark fringes occur is determined by the relationship between the path difference and the wavelength:
  • \( d \sin \theta = (m + 0.5) \lambda \)
Here, \( d \) is the distance between the two slits, and \( \theta \) is the angle with respect to the normal of the slits.
For every successive dark fringe, the order \( m \) increases by one. The angle of each fringe increases, resulting in the distribution of dark spots at various points on the screen.
This equation allows you to calculate the specific angle for any dark fringe based on its order and the arrangement of the slits.
Wavelength and Fringe Order
The concept of wavelength and fringe order helps in locating the precise positions of dark fringes. Wavelength \( \lambda \) is the distance between successive peaks of a wave, crucial for calculating interference patterns:
  • The wavelength determines how far apart the fringes will appear.
  • Fringe order \( m \) is essential to distinguish between different dark fringes.
For a given fringe order, the path difference is set to \( (m+0.5) \lambda \), making the calculation straightforward when using the formula \( d \sin \theta = (m + 0.5) \lambda \).
This helps in finding the angle for any specific dark fringe and is vital for both theoretical predictions and practical applications in experiments.

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Most popular questions from this chapter

Violet light (wavelength \(=410 \mathrm{~nm}\) ) and red light (wavelength \(=660 \mathrm{~nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines \(/ \mathrm{cm}\). This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the third-order maximum, (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

In a single-slit diffraction pattern on a flat screen, the central bright fringe is \(1.2 \mathrm{~cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{~m}\). When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of \(1.9 \mathrm{~cm}\). What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\).

Two in-phase sources of waves are separated by a distance of \(4.00 \mathrm{~m}\). These sources produce identical waves that have a wave length of \(5.00 \mathrm{~m}\). On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located?

Concept Questions (a) What, if any, phase change occurs when light, traveling in air, reflects from the interface between the air and a soap film \((n=1.33) ?\) (b) What, if any, phase change occurs when light, traveling in a soap film, reflects from the interface between the soap film and a glass plate \((n=1.52) ?(\mathrm{c})\) Is the wavelength of the light in a soap film greater than, smaller than, or equal to the wavelength in a vacuum? Problem A soap film \((n=1.33)\) is \(465 \mathrm{nm}\) thick and lies on a glass plate \((n=1.52)\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm}\), travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

Concept Questions (a) In a single-slit diffraction pattern the width of the central bright fringe is defined by the location of the first dark fringe that lies on either side of it. For a given slit width, does the width of the central bright fringe increase, decrease, or remain the same as the wavelength of the light increases? (b) For a given wavelength, does the width of the central bright fringe increase, decrease, or remain the same as the slit width increases? (c) When both the wavelength and the slit width change, it is possible for the width of the central bright fringe to remain the same. What condition must be satisfied for this to happen? In each case, give your reasoning. Problem A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{~m}\). When light with a wavelength of \(\lambda_{1}=510\) tim passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ) and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\).
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