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Karnataka Board PUCPUC Science Class 11

In a Double Slit Interference Experiment, the Separation Between the Slits is 1.0 Mm, the Wavelength of Light Used is 5.0 × 10−7 M and the Distance of the Screen from the Slits is 1.0m. - Physics

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Question

In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?

Sum

Solution

Given

Separation between the two slits,

\[d = 1  mm =  {10}^{- 3}   m\]

Wavelength of the light used,

\[\lambda = 5 . 0 \times  {10}^{- 7}   m\]

Distance between screen and slit,

\[D = 1  m\]

(a) The distance of the centre of the first minimum from the centre of the central maximum, \[x = \frac{\text{width of central maxima}}{2}\]

That is,

\[x = \frac{\beta}{2} = \frac{\lambda D}{2d}...........(1)\]

\[= \frac{5 \times {10}^{- 7} \times 1}{2 \times {10}^{- 3}}\]

\[   = 2 . 5 \times  {10}^{- 4}   m = 0 . 25  mm\]

(b) From equation (1),

fringe width,

\[\beta = 2 \times x = 0 . 50  mm\]

So, number of bright fringes formed in one centimetre (10 mm) = \[\frac{10}{0 . 50} = 20\]

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Chapter 17: Light Waves - Exercise [Page 381]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 17 Light Waves
Exercise | Q 7 | Page 381

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