Question

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?

Answer 1

Answer:

$5.86\times 10^{-7}\ \text{m}$

Explanation:

d = Slit separation = 0.11 mm

$\theta$ = Angle = $0.61^{\circ}$

m = Order = 2

$\lambda$ = Wavelength

We have the relation

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

The wavelength of the light is $5.86\times 10^{-7}\ \text{m}$.