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

`5.86* 10^(-7)\ \text{m}`

Step-by-step 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=(d\sin\theta)/(m)\\\Rightarrow \lambda=(0.11* 10^(-3)* \sin0.61^(\circ))/(2)\\\Rightarrow \lambda=5.86* 10^(-7)\ \text{m}`

The wavelength of the light is
`5.86* 10^(-7)\ \text{m}`.