In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?

  • Option 1)

    0.5 mm

  • Option 2)

    0.02 mm

  • Option 3)

    0.2 mm

  • Option 4)

    0.1 mm

 

Answers (1)

We need to obtain 10 maxima of double slit with in central maxima of single slit

For single slit, angular width of central maxima = \frac{2\lambda }{b}

or \frac{y}{d}=\frac{2\lambda }{b}or y=\frac{2\lambda D}{b}----------------------(1)

For interference pattern

Fringe width = \frac{\lambda D}{d}---------------------(2)

According to question

\frac{2\lambda D}{b}=10.\frac{\lambda D}{d}

\therefore b=\frac{d}{5}=0.2mm

 

 


Option 1)

0.5 mm

This option is incorrect

Option 2)

0.02 mm

This option is incorrect

Option 3)

0.2 mm

This option is correct

Option 4)

0.1 mm

This option is incorrect

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