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  • Need explanation for: Interference was observed in interference chamber when air is present. Now the chamber is evacuated and if the same light is used, then for the same arrangement:

Interference was observed in interference chamber when air is present. Now the chamber is evacuated and if the same light is used, then for the same arrangement:

  • Option 1)

    No interference pattern will be observed

  • Option 2)

    Exactly same interference pattern will be obtained with better contract

  • Option 3)

    The fringe width is slightly decreased

  • Option 4)

    The fringe width is slightly increased

 

Answers (1)

 

Fringe Width -

\beta = \frac{\lambda D}{d}
 

- wherein

\beta = y_{n+1}-y_{n}

y_{n+1}= Distance of\left ( n+1 \right )^{th}

Maxima= \left ( n+1 \right )\frac{\lambda D}{d}

y_{n}=Distance of n^{th}

 maxima = \frac{n\lambda D}{d}

 

 In young's interference fring width

\beta =\frac{\lambda D}{d}

when the medium change from air to vacuum, \lambda will increase and hence fringe width will increase


Option 1)

No interference pattern will be observed

Incorrect

Option 2)

Exactly same interference pattern will be obtained with better contract

Incorrect

Option 3)

The fringe width is slightly decreased

Incorrect

Option 4)

The fringe width is slightly increased

Correct

Posted by

Vakul

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