Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, The maximum number of possible interference maxima for slit-­separation equal to twice the wavelength in Young’s double­slit experiment is

The maximum number of possible interference  maxima for slit-­separation equal to twice the wavelength in Young’s double­slit experiment is

  • Option 1)

    infinite

  • Option 2)

    five

  • Option 3)

    three

  • Option 4)

    zero.

 

Answers (1)

As we learnt in

Fraunhofer Diffraction -

b\sin \theta = n\lambda
 

- wherein

Condition of nth minima.

b= slit width

\theta = angle of deviation

 

 

 

For interference maxima  d\sin \Theta = n\lambda

\therefore \: \: 2\lambda \sin \Theta = n\lambda

or= \sin \Theta = n/2

This equation is satisfied if n= -2,-1,0,1,2 \sin \Theta is never greater than (+1), less than (-1)

Maximum number of maxima can be five.

Correct option is 2.


Option 1)

infinite

This is an incorrect option.

Option 2)

five

This is the correct option.

Option 3)

three

This is an incorrect option.

Option 4)

zero.

This is an incorrect option.

Posted by

Sabhrant Ambastha

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

How to crack JEE Mains 2025 in 1 month? Follow study plan, revise high-score topics, experts advice
anam-khan

IIT JEE aspirant from Bihar found hanging in Kota hostel; 17th suicide this year
anam-khan

JEE Main 2025 just a month away; section-wise preparation tips, important topics, strategies

Latest Question