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 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?

Answer 1

Given that: $D=2mm=2\times {10}^{-3}ml=600nm=6\times {10}^{-7}m,I_{max}=0.20W/m^2,D=2,$
For the point, y = 0.5 cm
We known path difference $=x=\frac{yd}{D}$
$=0.5\times {10}^{-2}\times 2\times \frac{{10}^{-3}}{2}=5\times {10}^{-6}m$
So, the correspoinding phase difference is ,
$\varphi =\frac{2\pi x}{\lambda }=\frac{2\pi \times 5\times {10}^{-6}}{6\times {10}^{-7}}=\frac{50\pi }{3}=\frac{3\times 16\pi +2\pi }{3}\Rightarrow \varphi =\frac{2\pi }{3}$
So, the amplitude of the resulting wave at the point y = 0.5 cm is,
$A=\sqrt{r^2+r^2+2r^2\cos \left(\frac{2\pi }{3}\right)}=\sqrt{r^2+r^2-r^2}=rSince,\frac{I}{I_{max}}=\frac{A^2}{\left(2r^2\right)}$
[since, maximum amplitude - 2r]
$\Rightarrow \frac{I}{0.2}=\frac{A^2}{4r^2}=\frac{r^2}{4r^2}\Rightarrow I=\frac{0.2}{4}=0.05$