Question

In a fraunhofer diffraction by single slit of width d with incident light of wavelength \(5500 \AA\) the first minimum is observed at angle of \(30^{\circ}\). The first secondary maximum is observed at an angle \(\theta=\) (A) \(\sin ^{-1}(1 / \sqrt{2})\) (B) \(\sin ^{-1}(3 / 4)\) (C) \(\sin ^{-1}(\sqrt{3} / 2)\) (D) \(\sin ^{-1}(1 / 4)\)

Short answer

The angle for the first secondary maximum is given by: \( \theta_{max} = \sin^{-1}\left(\left(\frac{3}{2}\right)\frac{5500 × 1e-10}{d}\right) \) After substituting λ and d values and comparing to the given options, the correct answer is (B) \( \sin^{-1}(3 / 4) \).

Step-by-step solution

Convert the wavelength to meters

First, we need to convert the given wavelength of 5500 Å (which is in Angstroms) to meters. The conversion factor is 1 Å = 1e-10 meters. Therefore, the wavelength λ in meters is: λ = 5500 × 1e-10 m

Apply the Fraunhofer Diffraction Formula

For a single slit of width d, the angular position of the minima m in the Fraunhofer diffraction pattern is given by the formula: \( \sin \theta_{m} = \frac{m\lambda}{d} \) where m = 1, 2, 3... and θ is the angle between the central maximum and the minimum. We are given that the first minimum (m=1) is observed at an angle of 30°: \( \sin \theta_{1} = \frac{1\lambda}{d} \) Now, we can plug in the given angle and calculate the width of the slit.

Calculate the Width of the Slit

We know that the first minimum (m=1) appears at an angle of θ = 30°. Thus, \( \sin(30^{\circ}) = \frac{1\lambda}{d} \) Solve for d: \( d = \frac{1\lambda}{\sin(30^{\circ})} \) Now, substitute the value of λ (converted to meters): \( d = \frac{1(5500 × 1e-10)}{\sin(30^{\circ})} \) Calculate the width of the slit, d.

Find the Angle for the First Secondary Maximum

The condition for the secondary maxima is given by the formula: \( \sin \theta_{max} = \left(\frac{2k+1}{2}\right)\frac{\lambda}{d} \) where k = 1, 2, 3... and represents the order of the secondary maxima. For the first secondary maximum, k = 1: \( \sin \theta_{max} = \left(\frac{2(1)+1}{2}\right)\frac{\lambda}{d} \) Now substitute the values of λ and d to find the angle: \( \sin \theta_{max} = \left(\frac{3}{2}\right)\frac{5500 × 1e-10}{d} \) Therefore, θ_max is given by: \( \theta_{max} = \sin^{-1}\left(\left(\frac{3}{2}\right)\frac{5500 × 1e-10}{d}\right) \) Finally, compare the value obtained for θ_max to the given options (A), (B), (C), and (D) to find the correct answer.

Key concepts

Single Slit Diffraction

Single slit diffraction occurs when light waves pass through a narrow slit and spread out, creating a pattern of light and dark bands. This is known as the diffraction pattern. The phenomenon is due to the wave nature of light, where light waves bend around the edges of the slit, interfering with each other.

The diffraction pattern consists of a central bright spot, or maximum, flanked by alternating dark and bright bands. The dark bands are the minima, where destructive interference occurs, canceling out the light. These minima can be mathematically described by the formula:

\( \sin \theta_m = \frac{m\lambda}{d} \)

Here, \(m\) is the order of the minimum (\(m=1, 2, 3, ...\)), \(\lambda\) is the wavelength of the light, and \(d\) is the width of the slit. The angle \(\theta_m\) is the angle at which the minima occur. Understanding this pattern is crucial in the study of Fraunhofer diffraction, and it helps in determining various properties of the light and the slit.

Wavelength Conversion

In the context of physics and optics, precise measurement units are vital. Wavelength is commonly given in Angstroms (Å), where 1 Å is equal to \(1 \times 10^{-10}\) meters. To perform calculations in these units, it's essential to convert wavelengths from Angstroms to meters.

For example, a wavelength of 5500 Å becomes \(5500 \times 1e-10\) meters. This conversion is a crucial step in problems involving wave interference or diffraction, ensuring that all the values used are in consistent units.

Such consistency not only aids in calculation but also helps clearly visualize and understand the physical phenomena, as many scientific formulas, including the diffraction equations, require the use of base SI units.

Secondary Maximum

A secondary maximum in the diffraction pattern refers to a point, other than the central bright spot, where constructive interference of light waves occurs, leading to a smaller bright fringe. Secondary maxima are less intense than the central maximum because the light waves are not perfectly in phase as they add up, due to the geometry of wave diffraction through a slit.

The angular position \(\theta_{max}\) of these secondary maxima can be calculated using the formula:

\( \sin \theta_{max} = \left(\frac{2k+1}{2}\right)\frac{\lambda}{d} \)

Here, \(k\) denotes the order of the secondary maximum (\(k = 1, 2, 3, ...\)), indicating how far the maximum is from the central spot.

This pattern is intrinsic to any wave passing through a slit, providing key insights into the nature of light and the practical applications of diffraction patterns, such as in spectroscopy and optical engineering.