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Chapter 18: Problem 156

What will be the angular width of central maximum in Fraunhofer diffraction when light of wavelength \(6000 \AA\) is used and slit width is \(12 \times 10^{-5} \mathrm{~cm} ?\)

Short Answer

Expert verified
The angular width of the central maximum in Fraunhofer diffraction when the light has a wavelength of \(6000 \AA\) and the slit width is \(12 \times 10^{-5} \mathrm{~cm}\) is \(10 \: \text{radians}.\)

Step by step solution

01

Understand the Fraunhofer Diffraction equation for angular width

For Fraunhofer diffraction, the angular width of the central maximum can be found using the formula: \( \Delta\theta = \dfrac{2\lambda}{a} \) where: - \(\Delta\theta\) is the angular width of the central maximum - \(\lambda\) is the wavelength of the light - \(a\) is the slit width
02

Convert the given values to appropriate units

We are given the wavelength of the light in 脜ngstr枚ms and the slit width in centimeters: - Wavelength (位) = \(6000 \AA\) - Slit width (a) = \(12 \times 10^{-5}\) cm To perform calculations, we need to convert these values to appropriate units, i.e., meters: - 1 脜 = \(1 \times 10^{-10}\) m - 1 cm = \(1 \times 10^{-2}\) m Therefore: - Wavelength (位) = \(6000 \AA \times 10^{-10}~m/\AA = 6 \times 10^{-7} m\) - Slit width (a) = \((12 \times 10^{-5})~cm \times 10^{-2}~m/cm = 1.2 \times 10^{-6} m\)
03

Calculate the angular width of the central maximum

Now we can use the formula for the angular width in Fraunhofer diffraction and the converted values of the wavelength and slit width to find the angular width of the central maximum: \( \Delta\theta = \dfrac{2\lambda}{a} \) \( \Delta\theta = \dfrac{2 \times (6 \times 10^{-7}~m)}{1.2 \times 10^{-6}~m} \) \( \Delta\theta = \dfrac{12 \times 10^{-7}~m}{1.2 \times 10^{-6}~m} \) \( \Delta\theta = 10 \) Since the angular width is in radians, the angular width of the central maximum is 10 radians.

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Key Concepts

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

Angular Width
Angular width in the context of Fraunhofer diffraction represents the spread of the light's central maximum on either side of the central axis.
It is a crucial parameter in understanding diffraction patterns since it describes how wide the central bright region spans.
  • The formula for calculating angular width in Fraunhofer diffraction is \( \Delta\theta = \dfrac{2\lambda}{a} \).
  • This width is measured in radians, providing an angular measure of the diffraction pattern.
  • A larger angular width implies a wider central maximum, affecting the overall diffraction pattern.
This concept is essential in analyzing and predicting how different factors, such as wavelength and slit width, affect diffraction outcomes.
Central Maximum
The central maximum in a diffraction pattern is the bright area at the center, and it is the most intense part of the pattern.
In Fraunhofer diffraction, this occurs when the light waves constructively interfere.
  • Being the brightest and widest region, the central maximum provides valuable insight into the light's behavior when passed through a slit.
  • The width of this maximum is directly related to the wavelength of light and the slit width, which can be calculated using the angular width formula.
  • Experimental observations often focus on the central maximum, as its properties are integral to understanding the overall diffraction patterns.
Understanding the central maximum is critical because it dictates the sharpness and clarity of the diffraction image.
Wavelength
Wavelength, denoted as \( \lambda \), is the distance between consecutive crests or troughs of a wave.
In the context of Fraunhofer diffraction, the wavelength of the light source plays a significant role.
  • A longer wavelength results in a larger angular width of the central maximum, leading to a broader diffraction pattern.
  • Conversely, a shorter wavelength results in a narrower central maximum.
  • In the provided problem, the wavelength is given in 脜ngstr枚ms and must be converted to meters for calculations.
Recognizing the impact of wavelength on diffraction helps in accurately predicting and interpreting optical phenomena.
Slit Width
Slit width, represented as \( a \), is the physical opening through which light passes before creating a diffraction pattern.
This parameter significantly influences the pattern's appearance.
  • A narrower slit width results in a wider central maximum since the path differences allow greater spread of light.
  • A wider slit leads to a narrower central maximum, concentrating the light into a smaller area.
  • In the problem at hand, slit width is provided in centimeters, necessitating a conversion to meters for proper calculation.
Understanding slit width's influence is vital for controlling the resolution and visibility of diffraction patterns in practical applications.

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Most popular questions from this chapter

Focal length of an equiconvex lens is \(20 \mathrm{~cm}\). If we cut it once perpendicular to principle axis, and then along principal axis, then focal length of each part will be (A) \(20 \mathrm{~cm}\) (B) \(10 \mathrm{~cm}\) (C) \(40 \mathrm{~cm}\) (D) \(5 \mathrm{~cm}\)

Two pointed white dots are \(1 \mathrm{~mm}\) apart on a block paper. They are viewed by eye of pupil diameter \(3 \mathrm{~mm}\) approximately. What is the maximum distance at which these dots can be resolved by the eyes? [Take wavelength of light \(=500 \mathrm{~nm}]\) [2005] (A) \(1 \mathrm{~m}\) (B) \(5 \mathrm{~m}\) (C) \(3 \mathrm{~m}\) (D) \(6 \mathrm{~m}\)

A simple telescope, consisting of an objective of focal length \(30 \mathrm{~cm}\) and a single eye lens of focal length \(6 \mathrm{~cm}\). Eye observes final image in relaxed condition. If the angle subtended on objective is \(1.5^{\circ}\) then image will subtend angle of (A) \(6.5^{\circ}\) (B) \(7.5^{\circ}\) (C) \(0.3^{\circ}\) (D) \(2.5^{\circ}\)

Two convex lenses placed in contact form the image of a distant object at \(P\). If the lens \(B\) is moved to the right, the image will (A) move to the left. (B) move to the right. (C) remain at \(P\). (D) move either to the left or right, depending upon focal lengths of the lenses.

A point object is moving along the principal axis of a concave mirror at rest of focal length \(30 \mathrm{~cm}\) with speed \(5 \mathrm{~m} / \mathrm{s}\) towards the mirror. Find the speed of image of object when object is at a distance \(60 \mathrm{~cm}\) from mirror.
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