Respuesta :
Solution :
The expression for the intensity of light is given by :
[tex]$I=I_{max}\left(\frac{\sin \frac{\pi a \sin \theta}{\lambda}}{\frac{\pi a \sin \theta}{\lambda} }\right)^2$[/tex]
For a small angle, θ
sin θ = tan θ
[tex]$=\frac{y}{L}$[/tex]
Therefore the above equation becomes,
[tex]$I=I_{max}\left(\frac{\sin \frac{\pi a y}{\lambda L}}{\frac{\pi a y}{\lambda L} }\right)^2 $[/tex]
The given data is
λ = 546.1 nm
L = distance between the slit and the screen = 140 cm
= 1.40 m
a = width of the slit
= [tex]$0.50 \times 10^{-3} \ m$[/tex]
Therefore,
[tex]$I=I_{max}\left(\frac{\sin \frac{\pi \times 0.50 \times 10^{-3} \times 4.10 \times 10^{-3}}{546.1 \times 10^{-9} \times 1.20}}{\frac{\pi \times 0.50 \times 10^{-3} \times 4.10 \times 10^{-3}}{546.1 \times 10^{-9} \times 1.20} }\right)^2 $[/tex]
[tex]$=\left(\frac{0.170}{9.82}\right)^2$[/tex]
[tex]$= 2.89 \times 10^{-4} \ I_{max}$[/tex]
Therefore the fractional intensity is [tex]$\frac{I}{I_{max}}= 2.89 \times 10^{-4} $[/tex]
