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  • Solve! In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

  • Option 1)

    Unchanged

  • Option 2)

    Halved

  • Option 3)

    Doubled

  • Option 4)

    Quadrupled

 

Answers (1)

best_answer

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

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}

 

 \beta = \frac{\lambda D}{d}

\therefore D_{2} =2D_{1}

     d_{2} =\frac{d_{1}}{2}

\therefore \frac{\beta _{2}}{\beta _{1}} = \left ( \frac{D_{2}}{D_{1}} \right )\left ( \frac{d_{1}}{d_{2}} \right ) = 2\times 2 = 4

fringe width is quadrupled.


Option 1)

Unchanged

Incorrect

Option 2)

Halved

Incorrect

Option 3)

Doubled

Incorrect

Option 4)

Quadrupled

Correct

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