contestada

In a double-slit experiment, the second-order bright fringe is observed at an angle of 0.61°. If the slit separation is 0.11 mm, then what is the wavelength of the light?

Respuesta :

Answer:

[tex]5.86\times 10^{-7}\ \text{m}[/tex]

Explanation:

d = Slit separation = 0.11 mm

[tex]\theta[/tex] = Angle = [tex]0.61^{\circ}[/tex]

m = Order = 2

[tex]\lambda[/tex] = Wavelength

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{d\sin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.11\times 10^{-3}\times \sin0.61^{\circ}}{2}\\\Rightarrow \lambda=5.86\times 10^{-7}\ \text{m}[/tex]

The wavelength of the light is [tex]5.86\times 10^{-7}\ \text{m}[/tex].