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 In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used.  A person standing near the slits is looking at the fringe pattern.  When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits.  If the angular resolution of the eye is  
\frac{1^{\circ}}{60{}'}  the value of d0 is close to :

  • Option 1)

     1 mm

  • Option 2)

    2 mm

  • Option 3)

    4 mm

  • Option 4)

     3 mm

 

Answers (1)

As we learnt in

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}

 

 For a particular distance d0 between the slits, the eye is not able to resolve two consecutive bright fringes.

Now \theta=\frac{\beta}{D}  but  \beta=\frac{\lambda}{d_{0}}

\therefore\ \; \theta=\frac{\lambda d_{0}}{d_{0}}=\frac{600\times 10^{-9}m}{\frac{1}{60}\times \frac{\pi}{180}rad}=2.06\times10^{-3}m \simeq2mm

Correct option is 2.

 


Option 1)

 1 mm

This is an incorrect option.

Option 2)

2 mm

This is the correct option.

Option 3)

4 mm

This is an incorrect option.

Option 4)

 3 mm

This is an incorrect option.

Posted by

Vakul

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