Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 2304

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

Expert verified
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

01

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
02

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.
03

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.
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two thin lenses of focal length \(\mathrm{f}_{1}\) and \(\mathrm{f}_{2}\) are coaxially placed in contact with each other then the power of combination is (A) \(\left[\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right) / 2\right]\) (B) \(\sqrt{\left(f_{1} / f_{2}\right)}\) (C) $\left[\left(\mathrm{f}_{1} \mathrm{f}_{2}\right) /\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\right]$ (D) $\left[\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right) /\left(\mathrm{f}_{1} \mathrm{f}_{2}\right)\right]$

For a prism of refractive index \(\sqrt{3}\), the angle of minimum deviation is equitation is equal to the angle of prism, then angle of the prism is (A) \(60^{\circ}\) (B) \(90^{\circ}\) (C) \(45^{\circ}\) (D) \(180^{\circ}\)

Wave light travels from an optically rarer medium to an optically denser medium its velocity decrease because of change in (A) frequency (B) wavelength (C) amplitude (D) phase

A ray of light passes through a prism having refractive index \((\mathrm{n}=\sqrt{2})\), Suffers minimum deviation If angle of incident is double the angle of refraction within prism then angle of prism is (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)

An air bubble inside glass slab \((\mathrm{n}=1.5)\) appear from one side at $6 \mathrm{~cm}\( and from other side at \)4 \mathrm{~cm}$. Then the thickness of glass slab is \(\mathrm{cm}\) (A) 5 (B) 10 (C) 15 (D) 20
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Email

Read Explanation

Linguistic Terms

Read Explanation

Phonology

Read Explanation

Pragmatics

Read Explanation

Syntax

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free