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What will be the angular width of central maximum in Fraunhofer diffraction when light of wavelength \(6000 \mathrm{~A}\) 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 is approximately \(0.0573°\).

Step by step solution

01

Convert given quantities into same units

First, we need to convert both the wavelength and the slit width into the same units. Since the given wavelength is in Ångströms (Å) and the slit width is in centimeters (cm), we will convert them both to meters (m) as it is the SI unit. Given: Wavelength of light: \(λ = 6000 Å = 6000 × 10^{-10} m\) Slit width: \(a = 12 × 10^{-5} cm = 12 × 10^{-5} × 10^{-2} m = 12 × 10^{-7} m\)
02

Apply the formula for angular width

Now that we have the wavelength and slit width in meters, we can apply the formula for angular width in Fraunhofer diffraction: \(Δθ = \dfrac{2λ}{a}\)
03

Substitute the values and calculate the angular width

Next, we substitute the given values into the formula for angular width and calculate the result: \( Δθ = \dfrac{2 × 6000 × 10^{-10}}{12 × 10^{-7}} = \dfrac{12000 × 10^{-10}}{12 × 10^{-7}} = \dfrac{12000}{12} × \dfrac{10^{-10}}{10^{-7}} = 1000 × 10^{-3} = 1.0 \times 10^{-3} \, rad \)
04

Convert radians to degrees

Now that we have the angular width in radians, we can convert it to degrees: \(1° = \dfrac{180°}{π \, rad}\) \( Δθ = 1.0 × 10^{-3} \, rad × \dfrac{180°}{π \, rad} ≈ 0.0573° \) The angular width of the central maximum in Fraunhofer diffraction is approximately 0.0573°.

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Key Concepts

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

Angular Width
In the context of Fraunhofer diffraction, the angular width is a measure of the size of the pattern created when light passes through a slit and strikes a screen. This pattern consists of a central bright region known as the central maximum, flanked by alternating dark and bright bands. The angular width is specifically the angle
  • between the first minimum on one side of the central maximum
  • to the first minimum on the other side
To compute the angular width in a situation involving diffraction, we use the formula:\[Δθ = \dfrac{2λ}{a}\]Here:
  • \(Δθ\) represents the angular width
  • \(λ\) is the wavelength of the light used
  • \(a\) is the width of the slit.
Being clear on these terms aids in precisely calculating the diffraction pattern and appreciating how slight changes in wavelength or slit width can alter outcomes drastically.
Wavelength Conversion
Wavelength conversion is a crucial step when dealing with problems in optics, especially those involving measurements in different unit systems. The wavelength of light often needs to be converted from Ångströms (Å) to meters (m) for analysis in a consistent unit system like the International System of Units (SI).
1 Ångström is equivalent to \(10^{-10}\) meters, so to convert a wavelength from Ångströms to meters, multiply by \(10^{-10}\). For instance, in the original exercise, we were given:
  • Wavelength \(λ = 6000 Ã…\) which we converted to meters as: \(6000 × 10^{-10} = 6.0 × 10^{-7} m\).
Similarly, ensure that quantities such as slit width are similarly converted to meters.
This conversion facilitates accuracy in subsequent calculations for phenomena like angular width and central maximum in diffraction patterns.
Central Maximum Calculation
Central maximum calculation in Fraunhofer diffraction helps us understand the core features of the diffraction pattern when coherent light illuminates a slit. The central bright fringe, or central maximum, is the most intense part of the pattern. The calculation for locating the position of the central maximum is primarily done using the same variables as determining the angular width, but the emphasis is on understanding
  • how changes in slit width and wavelength affect its size and intensity.
The central maximum is pivotal because it offers a wealth of information about the light's interaction with the slit:
  • It behaves as a benchmark for understanding other minima and maxima in the diffraction pattern
  • It also indicates how closely spaced the maxima and minima are,
  • influenced by the slit dimensions and the wavelength used.
By mastering central maximum calculations, learners can predict and analyze diffraction behavior in real-world applications effectively.

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Most popular questions from this chapter

Consider the situation shown in the Fig. 18.51. Two slits \(S_{1}\) and \(S_{2}\) are placed symmetrically about the line \(O P\) which is perpendicular to screen and bisector to line joining the slits. The space between screen and slits is filled with a liquid of refractive index \(\mu_{3}, \mathrm{~A}\) plate of thickness \(t\) and refractive index \(\mu_{2}\) is placed in front of one of the slit. A source \(S\) is placed above \(O P\) at a distance \(d\) in front of slit. Given that \(D=1 \mathrm{~m}, d=2 \mathrm{~mm}, t=6 \times 10^{-6} \mathrm{~m}, \mu_{2}=1.2\), \(\mu_{3}=1.8\), Choose the correct alternatives(A) Position of central maximum from point \(P\) is \(2 \mathrm{~mm}\) (B) Position of central maximum from point \(P\) is \(1 \mathrm{~mm}\) (C) If slab is removed, the central maximum shift by a distance of \(2 \mathrm{~mm}\). (D) If slab is removed, the central maximum shift by a distance of \(1 \mathrm{~mm}\).

A thin convergent glass lens \(\left(\mu_{g}=1.5\right)\) has a power of \(+5.0 D .\) When this lens is immersed in a liquid of refractive index \(\mu_{i}\), it acts as a divergent lens of focal length \(100 \mathrm{~cm}\). The value of \(\mu_{l}\) is (A) \(4 / 3\) (B) \(5 / 3\) (C) \(5 / 4\) (D) \(6 / 5\)

In the diagram shown, the object is performing SHM according to the equation \(y=2 A \sin (\omega t)\) and the plane mirror is performing SHM according to the equation \(Y=-A \sin \left(\omega t-\frac{\pi}{3}\right) .\) The diagram shows the state of the object and the mirror at time \(t=0 \mathrm{~s}\). The minimum time from \(t=0 \mathrm{~s}\) after which the velocity of the image becomes equal to zero is(A) \(\frac{\pi}{3 \omega}\) (B) \(\frac{3 \pi}{\omega}\) (C) \(\frac{\pi}{6 \omega}\) (D) \(\frac{2 \pi}{3 \omega}\)

The principle of optical fibre is (A) Diffraction (B) Polarization (C) Interference (D) Total internal reflection

A circular beam of light of diameter \(d=2 \mathrm{~cm}\) falls on a plane surface of glass. The angle of incidence is \(60^{\circ}\) and refractive index of glass is \(\mu=3 / 2 .\) The diameter of the refracted beam is (A) \(4.0 \mathrm{~cm}\) (B) \(3.0 \mathrm{~cm}\) (C) \(3.26 \mathrm{~cm}\) (D) \(2.52 \mathrm{~cm}\)

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