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Ch 36: Diffraction
Chapter 35, Problem 36

Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

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Identify the given values: wavelength (λ) = 500.0 nm, distance to screen (L) = 90.0 cm, distance between bright bands (y) = 1.00 cm, and missing orders (m) = 3.
Convert all measurements to meters for consistency in units. For example, convert the wavelength from nanometers to meters by multiplying by $10^{-9}$, and convert distances from centimeters to meters by multiplying by $0.01$.
Use the formula for the position of the m-th bright fringe in a double-slit interference pattern, $y_m = m \frac{\lambda L}{d}$, where $d$ is the slit separation. Since the third order is missing, set up the equation for the third order and solve for $d$.
To find the slit width (a), use the condition that the minima occur at angles where $a \sin(\theta) = m\lambda$. Use the small angle approximation $\sin(\theta) \approx \tan(\theta) = \frac{y_m}{L}$, and solve for $a$ using the missing order condition.
Check the consistency of the results by substituting back the values of $a$ and $d$ into the equations for the interference and diffraction patterns to ensure they align with the given conditions of the problem.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Interference of Light

Interference occurs when two or more light waves overlap, resulting in a new wave pattern. In the context of a double-slit experiment, constructive interference produces bright bands (maxima) where the waves are in phase, while destructive interference leads to dark bands (minima) where the waves are out of phase. Understanding this principle is crucial for analyzing the resulting pattern from the slits.
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Young's Double-Slit Experiment

Young's double-slit experiment demonstrates the wave nature of light through the creation of an interference pattern. The distance between the slits and the screen, along with the wavelength of the light, determines the spacing of the bright and dark bands. The formula for the position of these bands is essential for calculating slit width and separation.
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Young's Double Slit Experiment

Wavelength and Slit Separation

The wavelength of light is a critical factor in determining the interference pattern produced by the slits. The separation between the slits affects the angle at which light waves interfere, influencing the distance between bright bands on the screen. By applying the relationship between wavelength, slit separation, and the observed pattern, one can derive the necessary dimensions of the slits.
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Related Practice
Textbook Question
A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation?
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Textbook Question
A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (b) What is the intensity at this point, if the intensity at the center of the central maximum is I0?
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Textbook Question
Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00x10^-4 W/m2, what is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?
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Textbook Question

Two satellites at an altitude of 1200 km are separated by 28 km. If they broadcast 3.6-cm microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh’s criterion) the two transmissions?

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