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# In a Young'S Double Slit Experiment, the Separation Between the Slits = 2.0 Mm, the Wavelength of the Light = 600 Nm and the Distance of the Screen from the Slits = 2.0 M. - Physics

Sum

In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W m−2, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes?

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#### Solution

Given:-

Separation between the slits,

$d = 2 mm = 2 \times {10}^{- 3} m$

Wavelength of the light,

$\lambda = 600 nm = 6 \times {10}^{-7} m$

Distance of the screen from the slits, D = 2.0 m

$I_\max = 0 . 20 W/ m^2$

For the point at a position $y = 0 . 5 cm = 0 . 5 \times {10}^{- 2} m,$

path difference, $∆ x = \frac{yd}{D}.$

$\Rightarrow ∆ x = \frac{0 . 5 \times {10}^{- 2} \times 2 \times {10}^{- 3}}{2}$

$= 5 \times {10}^{- 6} m$

So, the corresponding phase difference is given by

$∆ \phi = \frac{2\pi ∆ x}{\lambda} = \frac{2\pi \times 5 \times {10}^{- 6}}{6 \times {10}^{- 7}}$

$= \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$

or  $∆ \phi = \frac{2\pi}{3}$

So, the amplitude of the resulting wave at point y = 0.5 cm is given by

$A = \sqrt{a^2 + a^2 + 2 a^2 \cos \left( \frac{2\pi}{3} \right)}$

$= \sqrt{a^2 + a^2 - a^2} = a$

Similarly, the amplitude of the resulting wave at the centre is 2a.

Let the intensity of the resulting wave at point y = 0.5 cm be I.

Since $\frac{I}{I_\max} = \frac{A^2}{\left( 2a \right)^2},$ we have

$\frac{I}{0 . 2} = \frac{A^2}{4 a^2} = \frac{a^2}{4 a^2}$

$\Rightarrow I = \frac{0 . 2}{4} = 0 . 05 W/ m^2$

Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m2.

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 17 Light Waves
Q 29 | Page 382
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