Question

A thin beam of light of wavelength 50 μm (in the infrared portion of the em spectrum) goes through

Answer 1

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 :

$d=\frac{n \lambda D}{\Delta y}$

  $=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$

  = 0.000018203 m

  = 0.0182 mm

  = 0.018 mm

The intensity of light is given by :

$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$   , where $\beta=\frac{2 \pi D \sin \theta}{\lambda}$

$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

               = 0.1854

               ≈ 0.18

$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$

  $=I_0 \times 0.81$

  = 2  x0.81

  $= 1.62 \ W/m^2$