Question

A Young's double-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.48 mm. The interference pattern on the screen 3.7 m away shows the first maximum 5.1 mm from the center of the pattern. What is the wavelength of the light in nm

Answer 1

Wavelength of light used will be equal to 661 nm

Step-by-step explanation:

We have given distance between slits d = 0.48 mm =
`0.48* 10^(-3)m`

It is given interference pattern is 3.7 m away so D = 3.7 m

First maximum from the center is 5.1 mm

So
`y=5.1* 10^(-3)m`

Distance of the first maximum from the center is equal to
`y=(\lambda D)/(d)`

`5.1* 10^(-3)=(\lambda* 3.7)/(0.48* 10^(-3))=661* 10^(-9)=661nm`

So wavelength of light used will be equal to 661 nm