Explanation:
It is given that,
Slit width, [tex]d=5.7\times 10^{-4}\ m[/tex]
A diffraction pattern is formed on a flat screen located 4.0 m away, L = 4 m
The distance between the middle of the central bright fringe and the first dark fringe is 4.2 mm, y = 4.2 mm = 0.0042 m
Let [tex]\lambda[/tex] is the wavelength of the light.
Using condition of diffraction as,
[tex]d\ sin\ \theta=n\lambda[/tex]
[tex]\lambda=\dfrac{d\ sin\theta}{n}[/tex]..............(1)
Also, [tex]tan\theta=\dfrac{y}{L}[/tex]
[tex]\theta=tan^{-1}(\dfrac{y}{L})=tan^{-1}(\dfrac{0.0042}{4})=0.060[/tex]
[tex]\lambda=\dfrac{5.7\times 10^{-4}\ sin(0.060)}{1}[/tex]
[tex]\lambda=596\ nm[/tex]
So, the wavelength of the light is 596 nm. Hence, this is the required solution.
