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In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern

  • Option 1)

    The intensities of both the maxima and the minima increases

  • Option 2)

    The intensity of the maxima increases and the minima has zero intensity

  • Option 3)

    The intensity of maxima decreases and that of minima increases

  • Option 4)

    The intensity of maxima decreases and the minima has zero intensity

 

Answers (1)

best_answer

 

Maximum amplitude & Intensity -

When \theta = 0,2\pi ---2n\pi
 

- wherein

A_{max }= A_{1}+A_{2}

I_{max }= \left ( \sqrt{I_{1}}+\sqrt{I_{2}} \right )^{2}

 

 and

 

Minimum amplitude & Intensity -

\theta = \left ( 2n+1 \right )\pi
 

- wherein

A_{min }= A_{1}-A_{2}

I_{min }= \left ( \sqrt{I_{1}}-\sqrt{I_{2}} \right )^{2}

 

 In this case, since A \propto \ width of slits 

A_{1} = 2A

A_{2} = A

A^{max} = A_{1}+A_{2} = 3A and  A^{min} = \left ( A_{1} \right-A_{2} ) =A

Hence amplitude/intensity of both maxima and minima increases


Option 1)

The intensities of both the maxima and the minima increases

Correct

Option 2)

The intensity of the maxima increases and the minima has zero intensity

Incorrect

Option 3)

The intensity of maxima decreases and that of minima increases

Incorrect

Option 4)

The intensity of maxima decreases and the minima has zero intensity

Incorrect

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Plabita

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