Answer:
Wavelength, [tex]\lambda=1.28\times 10^{-6}\ m[/tex]
Explanation:
Given that,
Width of the slit, [tex]d=2.3\times 10^{-3}\ mm=2.3\times 10^{-6}\ m[/tex]
The angle between the first dark fringes on either side of the central maximum is 34 degrees
To find,
The wavelength of the light used.
Solution,
The equation of maxima is given by :
[tex]dsin\theta=n\lambda[/tex]
n = 1 here
[tex]\lambda=\dfrac{d\ sin\theta}{n}[/tex]
[tex]\lambda={d\ sin\theta}[/tex]
[tex]\lambda={2.3\times 10^{-6}\times sin(34)}[/tex]
[tex]\lambda=1.28\times 10^{-6}\ m[/tex]
So, the wavelength of the light used is [tex]1.28\times 10^{-6}\ m[/tex]. Hence, this is the required solution.
