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  • 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

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