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/m2.
