Respuesta :
Solution :
Given :
Wavelength of the thin beam of light, λ = 50 μm
Distance of the screen from the slit, D = 3.00 m
Width of the fringe, Δy = ±8.24 mm
Therefore, width of the slit is given by :
[tex]$d=\frac{n \lambda D}{\Delta y}$[/tex]
[tex]$=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$[/tex]
= 0.000018203 m
= 0.0182 mm
= 0.018 mm
The intensity of light is given by :
[tex]$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$[/tex] , where [tex]$\beta=\frac{2 \pi D \sin \theta}{\lambda}$[/tex]
[tex]$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$[/tex]
[tex]$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$[/tex]
Now, [tex]$\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$[/tex]
= 0.1854
≈ 0.18
[tex]$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$[/tex]
[tex]$=I_0 \times 0.81$[/tex]
= 2 x0.81
[tex]$= 1.62 \ W/m^2$[/tex]
