
A parallel beam of light of wavelength 4000Å passes through a slit of width 5×10-3m. What will be the angular spread of the central maxima in the diffraction pattern ?
A. 1.6×10-3 rad
B. 1.6×10-4 rad
C. 1.2×10-3 rad
D. 3.2×10-3 rad
E. 3.2×10-4 rad
Answer
164.1k+ views
Hint: As it is well known that central maxima is the brightest central portion of diffraction pattern and it has maximum intensity.It is between the 1st order minima from the centre of the screen on both sides of centre.
Formula used:
Here, formula will be:
nλ = asinθ , where n = nth fringe
λ = wavelength
a = slit width
θ = angular position of any point on the screen, taken from the slit centre
And tanθ = y/D where y = Position of minima
D = Separation between slit and source
Complete Step by Step Answer:
Firstly, we are required to understand that the angle subtended by the beam on the screen will be too small that we can take θ instead of sinθ and θ instead of tanθ.
So,nλ = aθ
Here, n = 1. Therefore, θ = /a
Also, θ = y/D
So, /a = y/D
y = λD/a
Hence, Width of central maximum = 2λD/a
Angular width of central maximum = 2θ = 2λ/a
Here, we are asked to find out angular width.
a = 5×10-3m, λ = 4000Å
So, $2 \times \dfrac{4000\times10^{-10}}{5\times10^{-3}}$ = 1600×10-7 rad = 1.6×10-4 rad
Hence, it is proved that option (B) is the correct option.
Note: It should remain in memory that it will be twice the λ/a because this width is on both sides of the screen due to which it becomes twice of θ. Unit conversion should be in standard form.
Formula used:
Here, formula will be:
nλ = asinθ , where n = nth fringe
λ = wavelength
a = slit width
θ = angular position of any point on the screen, taken from the slit centre
And tanθ = y/D where y = Position of minima
D = Separation between slit and source
Complete Step by Step Answer:
Firstly, we are required to understand that the angle subtended by the beam on the screen will be too small that we can take θ instead of sinθ and θ instead of tanθ.
So,nλ = aθ
Here, n = 1. Therefore, θ = /a
Also, θ = y/D
So, /a = y/D
y = λD/a
Hence, Width of central maximum = 2λD/a
Angular width of central maximum = 2θ = 2λ/a
Here, we are asked to find out angular width.
a = 5×10-3m, λ = 4000Å
So, $2 \times \dfrac{4000\times10^{-10}}{5\times10^{-3}}$ = 1600×10-7 rad = 1.6×10-4 rad
Hence, it is proved that option (B) is the correct option.
Note: It should remain in memory that it will be twice the λ/a because this width is on both sides of the screen due to which it becomes twice of θ. Unit conversion should be in standard form.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
