In the Young's double slit experiment the intensity produced by each one of the individual slits is

Question

In the Young's double slit experiment the intensity produced by each one of the individual slits is $I_{\mathrm{o}}$. The distance between two slits is 2 mm . The distance of screen from slits is 10 m . The wavelength of light is $6000 \AA$. The intensity of light on the screen in front of one of the slits is $\_\_\_\_$

Professor PC Thomas Classes · Answer

The position of the point on the screen exactly in front of one slit is $y = \frac{d}{2} = 1 	ext{ mm} = 10^{-3} 	ext{ m}$. The path difference at this point is $\Delta x = y \frac{d}{D} = (10^{-3}) \frac{2 	imes 10^{-3}}{10} = 2 	imes 10^{-7} 	ext{ m}$. The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6000 	imes 10^{-10}} 	imes 2 	imes 10^{-7} = \frac{2\pi}{6 	imes 10^{-7}} 	imes 2 	imes 10^{-7} = \frac{2\pi}{3} = 120^\circ$. The resultant intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi = I_0 + I_0 + 2I_0\cos(120^\circ) = 2I_0 + 2I_0(-0.5) = 2I_0 - I_0 = I_0$.