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Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

Short Answer

Expert verified
The wavelength of the light is 1.11 μm.

Step by step solution

01

Identify Given Information

From the problem, it is identified that the slit width (d) is 0.750mm or 0.750 x \(10^{-3}\) m, the distance between the screen and the slit (L) is 2.00m, and the distance to the first minimum (D) is 1.35mm or 1.35 x \(10^{-3}\) m.
02

Apply the formula for the wavelength of diffracted light

The wavelength of the light can be calculated using the formula for first minimum in a diffraction pattern from a single slit which is, \(\lambda = \frac{2Ld}{D}\). Plug in the values: \(\lambda = \frac{2 * 2* 0.750 * 10^{-3}}{1.35 * 10^{-3}}\).
03

Calculation and Answer

Compute the value to get the wavelength. Thus, \(\lambda = 1.11 * 10^{-6}\) m or 1.11 μm is the wavelength of the light.

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Key Concepts

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

Wavelength Calculation
Getting to grips with calculating the wavelength of light in a diffraction experiment is simpler than it might initially seem. Wavelength, often represented by the Greek letter \(\lambda\), is the distance between two consecutive peaks of a wave. In the context of optics, it's crucial for predicting how light behaves when encountering obstacles.
For single-slit diffraction, the wavelength can be calculated using the specific formula \[\lambda = \frac{Ld}{D}\]where \(L\) is the distance from the slit to the screen, \(d\) is the slit width, and \(D\) represents the distance from the central maximum to the first minimum on the screen.
This formula arises because the path length difference that causes the first minimum is effectively half a wavelength. When light waves emerge from different parts of the slit and interfere destructively, a diffraction pattern forms on the screen.
  • Remember to convert all measurements to meters to keep the units consistent.
  • Use precise measurements to ensure accurate wave predictions.
As demonstrated in the solved problem, plugging the respective values into the formula provides the wavelength, making it a straightforward but powerful calculation.
Single-Slit Diffraction
Single-slit diffraction is the bending of light waves as they pass through a narrow aperture, which leads to a distinctive pattern of light and dark bands. This phenomenon happens because different parts of the wavefront can interfere,producing patterns of constructive and destructive interference.Light traveling through a single slit does not continue in a straight line. Instead, it spreads out, creating a central bright fringe surrounded by darker and alternating lighter fringes on each side. This central position is known as the central maximum.
The points where the light is least intense, or the "minima" points, occur where waves interfere destructively - meaning they cancel each other out.
  • The condition for diffraction minima is given by the equation \(d \sin \theta = m \lambda\), where \(m\) is the order of the minimum (for example, \(m=1\) for the first minimum), \(d\) is the width of the slit,\(\theta\) is the angle to the minimum, and \(\lambda\) is the wavelength of the light.
  • This simple yet profound concept shows how even a seemingly tiny slit can significantly influence the shape and distribution of light beyond the aperture.
Understanding single-slit diffraction lays the foundation for more complex wave behavior studies in optics.
Optics
Optics is the branch of physics that focuses on the behavior and properties of light, and its interactions with matter. It encompasses various phenomena including reflection, refraction, and diffraction, the latter of which is a unique optical event. Diffraction, as discussed in this context, specifically refers to how waves bend around edges and through narrow openings. This behavior illustrates one of the fundamental properties of waves and can be observed not only in light, but in sound and water waves as well.
In optics, the study of diffraction provides essential insights into understanding wave behavior in various conditions. It ultimately helps in designing lenses for cameras, glasses, and telescopes, enabling them to manipulate light efficiently.
  • In everyday life, diffraction can be seen in the spreading of light waves, explaining why we can hear sound around corners and why stars appear to twinkle due to atmospheric effects.
  • The patterns created by diffraction are critical in technology industries, such as the manufacturing of optical equipment and developing devices for precise measurement.
The study of optics is instrumental in expanding our comprehension of wave phenomena and in the creation of technologically advanced optical devices.

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Most popular questions from this chapter

(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.

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

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

The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm} .\) White light falls at normal incidence on a diffraction grating that has 350 slits \(/ \mathrm{mm} .\) Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example \(36.4 .\) )

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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