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Coherent electromagnetic waves with wavelength \(\lambda=500 \mathrm{nm}\) pass through two identical slits. The width of each slit is \(a,\) and the distance between the centers of the slits is \(d=9.00 \mathrm{~mm}\). (a) What is the smallest possible width \(a\) of the slits if the \(m=3 \max -\) imum in the interference pattern is not present? (b) What is the next larger value of the slit width for which the \(m=3\) maximum is absent?

Short Answer

Expert verified
To make the m=3 maximum not present, the smallest possible width a of the slits is 1500nm. The next larger value of the slit width for which the m=3 maximum is absent is 2000nm.

Step by step solution

01

Calculate the smallest possible width a of the slits for abscent m=3 maximum

The formula for determining a when m=3 maximum is not present is \( a = m \cdot \lambda \). Plug in the given parameters. We take \(m = 3\) and wavelength \(\lambda = 500 \times 10^{-9} \mathrm{m}\). Calculate \(a = 3 \cdot 500 \times 10^{-9}\) to get the first possible width.
02

Calculate the next larger value for a where the m=3 maximum is abscent

To find the next larger value of a where the m=3 maximum is absent, we simply advance to the next higher order of m, which is \(m = 4\). The formula remains the same: \( a = m \cdot \lambda \). Plug in our new parameters. The calculation is \(a = 4 \cdot 500 \times 10^{-9}\). This gives us the next larger value for a.

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Key Concepts

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

Double-Slit Experiment
The double-slit experiment is a fascinating demonstration of wave interference. It was first performed by Thomas Young in the early 1800s. This simple yet powerful experiment shows how light and other waves interact when they pass through two neighboring openings or slits. Imagine shining a light, like sunlight filtered through a narrow slit, onto a barrier with two equally spaced openings. As the light passes through these openings, each slit acts as a source of waves. These waves then spread out, overlap, and create patterns on a screen placed behind the barrier. Why is this significant? - It demonstrates the wave nature of light. - It shows how waves can add together in some places to make a brighter spot (constructive interference) and cancel each other out in others, leading to darkness (destructive interference). - The overall pattern of light and dark bands on the screen is called an interference pattern, and it beautifully demonstrates how waves behave.
Wavelength
Wavelength, usually denoted by the Greek letter \( \lambda \), is a crucial concept in understanding wave mechanics, including the double-slit experiment. It represents the distance between consecutive crests or troughs in a wave. In simpler terms, it's the "length" of one complete cycle of a wave. Why does it matter? The wavelength of light affects the position and visibility of the interference pattern.Here's how it works:- Longer wavelengths lead to wider spacing between the bright and dark bands in an interference pattern. - Shorter wavelengths cause the bands to appear closer together.In the context of the double-slit experiment, using a wavelength of \(\lambda = 500 \text{ nm}\), as specified in the exercise, helps determine the conditions under which interference maxima (bright spots) occur or disappear.So, wavelength directly influences what we see when waves clash.
Interference Pattern
The interference pattern created in the double-slit experiment is a beautiful visual result of wave interference. When two light waves meet, they can either cooperatively add up (constructive interference) or cancel each other out (destructive interference).In a typical setup:- Bright bands on the screen are locations of constructive interference where the crests of one wave align with the crests of another.- Dark bands arise from destructive interference where the crest of one wave meets the trough of another and cancel each other out.The pattern you see—alternating bright and dark bands—is called the interference pattern.Factors affecting this pattern: - **Wavelength:** As mentioned, different wavelengths spread out the bands by different amounts. - **Distance between slits (\(d\)):** The greater the distance between the slits, the closer the bands are on the screen. - **Slit width (\(a\)):** This influences the sharpness and visibility of the bands. Understanding the interference pattern is key to grasping why certain maxima, like the \(m=3\) maximum in the exercise, might not appear.
Slit Width
The width of the slits, denoted by \(a\), plays a significant role in the double-slit experiment. It not only affects the sharpness of the interference pattern but also determines if certain maxima, like the \(m=3\) maximum from the exercise, will be absent.Why is slit width important?- If the slit width \(a\) is too narrow, different diffraction effects can blur the bands, making it hard to see clear interference patterns.- The formula \( a = m \cdot \lambda \) used in the exercise is derived to ensure that certain maxima are absent. For the given exercise:- With \(m=3\) and \(\lambda = 500 \text{ nm}\), the smallest slit width \(a\) ensuring the \(m=3\) maximum is absent is calculated as \( a = 3 \times 500 \times 10^{-9} \text{ m} \).- The next larger value is found by considering \(m=4\), resulting in \( a = 4 \times 500 \times 10^{-9} \text{ m} \).Understanding slit width helps determine how waves will interfere and what visible patterns will emerge on a screen.

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Most popular questions from this chapter

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is \(656.45 \mathrm{nm} ;\) for deuterium, the corresponding wavelength is \(656.27 \mathrm{nm}\). (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

Nearly monochromatic coherent light waves leave two rectangular slits in phase and at an angle of \(\theta=22.0^{\circ}\) with the normal. When the light reaches a distant screen, the waves from the center of one slit are \(344^{\circ}\) out of phase with the waves from the center of the other slit, and the waves from the top of either slit are \(172^{\circ}\) out of phase with the waves from the bottom of that slit. (a) How is the center-to-center distance between the slits related to the width of either slit? (b) Calculate the intensity at the screen for \(\theta=22.0^{\circ}\) if the intensity at \(\theta=0^{\circ}\) is \(0.234 \mathrm{~W} / \mathrm{m}^{2}\)

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits \(/ \mathrm{cm} ?\) (b) What is the second-order deviation of this wavelength? Assume normal incidence.

If you can read the bottom row of your doctor’s eye chart, your eye has a resolving power of 1 arcminute, equal to \(\frac{1}{60}\) degree. If this resolving power is diffraction limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume \(\lambda=550 \mathrm{nm}\).

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

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