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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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प्रश्न

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?

योग

उत्तर

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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अध्याय 17: Light Waves - Exercise [पृष्ठ ३८१]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 17 Light Waves
Exercise | Q 7 | पृष्ठ ३८१

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