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?

#### 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/m^{2}.