Question

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 center of the central maximum is $0.20W{m}^{-2}$, what will be this intensity at a point $0.5$ cm away from this center along the width of the fringes?

Answer 1

Path diff. = $x=\frac{yd}{\Delta }$

$=\frac{0.5\times {10}^{-2}\times 2\times {10}^{-3}}{2}$

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

Phase difference ,

$\varphi =\frac{2\pi x}{\lambda }$

$=\frac{2\pi \times 5\times {10}^{-6}}{6\times {10}^{-7}}$

$=\frac{50\pi }{3}$

$=10\pi +\frac{2\pi }{3}$

$=\frac{2\pi }{3}$

Amplitude ,

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

$=r$

$\frac{I}{I_{max}}=\frac{A^2}{2^2{r}^2}$

$\frac{I}{0.2}=\frac{A}{4r^2}=\frac{r^2}{4r^2}$

$I=\frac{0.2}{4}=0.05w/m^2$