(a) If the first-order maximum for monochromatic light falling on a double slit is at an angle of \(10.0^{\circ},\) at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest- order maximum possible here?

Short Answer

Expert verified
The angle for the second-order maximum is approximately \(19.47^{\circ}\), the angle for the first minimum is approximately \(5.74^{\circ}\), and there is no highest-order maximum possible for this given scenario.

Step by step solution

01

Write down the given information

The angle of the first-order maximum is given as \(10.0^{\circ}\).
02

Write down the formula for the maximum

The formula for the maximum in a double slit interference pattern is: \[d \sin{\theta} = m \lambda\]
03

Solve for the second-order maximum

We know the first-order maximum angle, so we can write the formula for that: \(d \sin{10.0^{\circ}} = 1 \cdot \lambda\) Now, we want to find the angle for the second-order maximum: \(d \sin{\theta_2} = 2 \cdot \lambda\) Divide the second equation by the first equation: \[\frac{\sin{\theta_2}}{\sin{10.0^{\circ}}} = 2\] Now, solve for \(\theta_2\): \[\theta_2 = \arcsin(2 \cdot \sin{10.0^{\circ}})\] Calculate the value: \[\theta_2 \approx 19.47^{\circ}\] So, the angle for the second-order maximum is approximately \(19.47^{\circ}\). #b) Finding the angle of the first minimum#
04

Write down the formula for the first minimum

The formula for the first minimum in a double slit interference pattern is: \[d \sin{\theta} = (m + \frac{1}{2}) \lambda\]
05

Solve for the first minimum angle

We want to find the angle for the first minimum, which corresponds to \(m = 0\): \(d \sin{\theta_1} = (0 + \frac{1}{2}) \cdot \lambda\) Divide the first minimum equation by the first-order maximum equation: \[\frac{\sin{\theta_1}}{\sin{10.0^{\circ}}} = \frac{1}{2}\] Now, solve for \(\theta_1\): \[\theta_1 = \arcsin(\frac{1}{2} \cdot \sin{10.0^{\circ}})\] Calculate the value: \[\theta_1 \approx 5.74^{\circ}\] So, the angle for the first minimum is approximately \(5.74^{\circ}\). #c) Finding the highest-order maximum possible#
06

Determine the condition for the highest-order maximum

The condition for the highest-order maximum is when the maximum and the minimum coincide: \[m \lambda = (m + \frac{1}{2}) \lambda\]
07

Solve for the highest-order maximum

Subtract the left side of the equation from the right side: \[\frac{1}{2} \lambda = m \lambda\] Divide both sides by \(\lambda\): \[m = \frac{1}{2}\] Since \(m\) has to be an integer, there is no highest-order maximum for this given scenario.

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Key Concepts

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

Interference Pattern
When light waves overlap, they create a phenomenon known as an 'interference pattern.' Imagine dropping two pebbles into a still pond at the same time. The ripples created by both pebbles interact with each other, producing areas where the water seems very still (destructive interference) and others where the ripples are higher (constructive interference). In wave optics, a similar effect occurs when light passes through two slits closely spaced apart - a setup known as Young's double-slit experiment.

The result is a series of light and dark bands on a screen behind the slits. These bands arise because the light waves from the two slits arrive at the screen either in phase, reinforcing each other for a bright band (constructive interference), or out of phase, canceling each other for a dark band (destructive interference). The position of these bands depends on the wavelength of the light and the distance between the slits, providing a visual map of how waves interfere with one another.
Wave Optics
Wave optics, or 'physical optics,' deals with the study of various phenomena that result from the wave nature of light, such as interference, diffraction, and polarization. Unlike geometric optics, where light is treated as rays that travel in straight lines, wave optics considers the wave characteristics of light. This allows for a more comprehensive understanding of the complex behaviors light exhibits when it encounters obstacles, openings, or other waves.

For instance, in the double-slit experiment we mentioned earlier, wave optics explains why light doesn't just project two slits onto a screen, but instead creates an intricate pattern of fringes. This is because the peaks and troughs of light waves interact in complex ways that are predictable through equations and principles derived from the study of wave optics.
Diffraction
Diffraction is the bending of waves around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. In the realm of light, diffraction is usually observed when a wavefront encounters a slit or edge. The degree of bending depends on the relative size of the wavelength of light to the size of the opening or obstacle.

In the double slit experiment, diffraction is the reason why light spreads out after passing through the slits and doesn't just move straight forward. This spreading allows the waves from the two slits to overlap and intermingle, resulting in a characteristic interference pattern. The narrower the slits, the greater the spread of the waves, and hence the more pronounced the interference. This concept is crucial because it links the wave nature of light to its observable effects in real-world scenarios.
Physics Education
Physics education aims to not only impart knowledge but also to develop critical thinking and problem-solving skills. When teaching concepts like double slit interference, creating relatable analogies such as the pebble-ripple example helps make abstract concepts more tangible. Simplifying complex equations and progressively building up to the complete solution can also aid understanding.

Including visual aids, like drawings of the interference pattern, and providing step-by-step solutions that clearly lay out the logic behind each step, greatly benefits students. They can follow the thought process and apply similar reasoning to other physics problems. Encouraging students to express the concepts in their own words, and asking them to predict outcomes before revealing results, can further deepen their grasp of the subject matter.

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Most popular questions from this chapter

What is the distance moved by the traveling mirror of a Michelson interferometer that corresponds to 1500 fringes passing by a point of the observation screen? Assume that the interferometer is illuminated with a 606 nm spectral line of krypton-86.

An intensity minimum is found for 450 nm light transmitted through a transparent film \((n=1.20)\) in air. (a) What is minimum thickness of the film? (b) If this wavelength is the longest for which the intensity minimum occurs, what are the next three lower values of \(\lambda\) for which this happens?

How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?

Two sources as in phase and emit waves with \(\lambda=0.42 \mathrm{m}\). Determine whether constructive or destructive interference occurs at points whose distances from the two sources are (a) 0.84 and \(0.42 \mathrm{m}\), (b) 0.21 and \(0.42 \mathrm{m},\) (c) 1.26 and \(0.42 \mathrm{m},\) (d) 1.87 and \(1.45 \mathrm{m},\) (e) 0.63 and \(0.84 \mathrm{m}\) and \((\mathrm{f}) 1.47\) and \(1.26 \mathrm{m}\).

Shown below is a double slit located a distance \(x\) from a screen, with the distance from the center of the screen given by \(y .\) When the distance \(d\) between the slits is relatively large, numerous bright spots appear, called fringes. Show that, for small angles (where \(\sin \theta \approx \theta,\) with \(\theta\) in radians), the distance between fringes is given by \(\Delta y=x \lambda / d\)

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