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What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at \(\theta=45.0^{\circ} ?\)

Short Answer

Expert verified
The ratio must be \( \sqrt{2} \).

Step by step solution

01

Understanding the Diffraction Condition

In single-slit diffraction, the minima occur at angles \( \theta \) where the path difference satisfies \( a \sin \theta = m \lambda \), with \( m \) being an integer (\( m = 1, 2, 3, \ldots \)) corresponding to the order of the minimum. For the first minimum, \( m = 1 \). We aim to find the ratio of the slit width \( a \) to the wavelength \( \lambda \).
02

Substitute Known Values for the First Minimum

We want to find the condition for the first minimum (\( m = 1 \)) at \( \theta = 45.0^{\circ} \). Therefore, substitute these values into the formula: \( a \sin(45^{\circ}) = \lambda \).
03

Calculate \( \sin(45^{\circ}) \)

The sine of \( 45^{\circ} \) is well known, being equal to \( \frac{\sqrt{2}}{2} \). Thus, we substitute this value into our expression for the first minimum.
04

Solve for the Ratio \( \frac{a}{\lambda} \)

Rearrange the expression to solve for \( \frac{a}{\lambda} \):\[ \frac{a}{\lambda} = \frac{1}{\sin(45^{\circ})} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}. \]
05

Conclusion

The ratio of the slit width \( a \) to the wavelength \( \lambda \) for the first diffraction minimum to occur at \( \theta=45.0^{\circ} \) is \( \sqrt{2} \).

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Key Concepts

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

Diffraction Minimum
The concept of a **diffraction minimum** is central in understanding single-slit diffraction patterns. When light waves encounter a slit, they spread out and create a pattern of alternating bright and dark spots. The dark spots in this pattern are known as diffraction minima, where destructive interference causes the light intensity to be at its lowest. The condition for these minima can be described by the formula:- **Path Difference:** \( a \sin \theta = m \lambda \) - Where: - \( a \) is the slit width. - \( \theta \) is the angle of diffraction. - \( m \) is the order of the minimum \((m = 1, 2, 3, \ldots)\). - \( \lambda \) is the wavelength of light.For the first diffraction minimum, \( m = 1 \), meaning that the smallest angle where complete darkness occurs is caused by a path difference equal to one wavelength. Understanding how these minima occur helps in predicting how light will behave when passing through any aperture, a key concept for many optical applications.
Wavelength
The **wavelength** of light is a measure of the distance between identical points in the wave cycle, usually from crest to crest. It is typically measured in meters, and the wavelength of visible light ranges from approximately 400 nm to 700 nm.Wavelength is a crucial factor in diffraction. It determines how much a light wave spreads after passing through a slit. Longer wavelengths will spread more than shorter ones when diffracted. In single-slit diffraction, altering the wavelength changes the position of the minima and maxima in the diffraction pattern:- **Shorter Wavelength**: - Results in closer minima and maxima. - Less spreading of the light.- **Longer Wavelength**: - Causes more spreading. - Minima and maxima will be further apart.The equation for the diffraction minima \( a \sin \theta = m \lambda \) directly links the wavelength with the angle \( \theta \) at which these minima occur. Thus, understanding wavelength is essential for predicting and utilizing diffraction patterns in optics.
Slit Width
The **slit width** \( a \) is another critical parameter in single-slit diffraction. This component influences how much the light waves spread when they pass through the slit. The width of the slit determines the diffraction angle \( \theta \) of light:- **Narrow Slit**: - Causes greater spreading and wider diffraction patterns. - Minima and maxima are spaced further apart.- **Wide Slit**: - Results in less spreading and more closely spaced diffraction patterns.The slit width is part of the diffraction condition equation: \( a \sin \theta = m \lambda \). This equation reveals that as the slit width decreases, the angle \( \theta \) increases, leading to the observation of a wider diffraction pattern for a given wavelength. Understanding how slit width affects the diffraction pattern is crucial, whether you're designing optical instruments or studying the properties of light. Changes in the slit width can dramatically alter the visibility and clarify the diffraction pattern, impacting the interpretation of the optical properties you're analyzing.

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Most popular questions from this chapter

A diffraction grating is made up of slits of width \(300 \mathrm{nm}\) with separation \(900 \mathrm{nm}\). The grating is illuminated by monochromatic plane waves of wavelength \(\lambda=600 \mathrm{nm}\) at normal incidence. (a) How many maxima are there in the full diffraction pattern? (b) What is the angular width of a spectral line observed in the first order if the grating has 1000 slits?

The wall of a large room is covered with acoustic tile in which small holes are drilled \(5.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm},\) and the wavelength of the room light to be \(550 \mathrm{nm} ?\)

In June \(1985,\) a laser beam was sent out from the Air Force Optical Station on Maui, Hawaii, and reflected back from the shuttle \(D i s-\) covery as it sped by \(354 \mathrm{~km}\) overhead. The diameter of the central maximum of the beam at the shuttle position was said to be \(9.1 \mathrm{~m}\) and the beam wavelength was \(500 \mathrm{nm}\). What is the effective diameter of the laser aperture at the Maui ground station? (Hint: A laser beam spreads only because of diffraction; assume a circular exit aperture.)

A diffraction grating having 180 lines/mm is illuminated with a light signal containing only two wavelengths, \(\lambda_{1}=400 \mathrm{nm}\) and \(\lambda_{2}=500 \mathrm{nm}\). The signal is incident perpendicularly on the grating. (a) What is the angular separation between the secondorder maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are superimposed? (c) What is the highest order for which maxima for both wavelengths are present in the diffraction pattern?

If you look at something \(40 \mathrm{~m}\) from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh's criterion? Assume the pupil of your eye has a diameter of \(4.00 \mathrm{~mm}\), and use \(500 \mathrm{nm}\) as the wavelength of the light reaching you.

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