Question

Coherent, monochromatic light is incident on a pair of slits that are 0.3 mm apart. When light passes through the slits and strikes a screen 1.5 m away, the distance between the 3rd and 5th order maximum is 5.5 mm. What is the wavelength of the light being used?

Answer 1

550 nm

Step-by-step explanation:

The formula for the double-slit diffraction experiment is:

`y=(m\lambda D)/(d)`

where

y is the distance of the m-th maximum from the central fringe

`\lambda` is the wavelength of the light used

D is the distance of the screen from the slits

d is the separation between the slits

In this problem we have:

`d=0.3 mm = 0.3\cdot 10^(-3) m` is the distance between the slits

`D=1.5 m` is the distance of the screen

We also know that the distance between the 3rd and 5th order maximum is 5.5 mm, which means that:

`y_5-y_3 = 5.5 mm = 5.5\cdot 10^(-3) m`

And we can rewrite this as:

`y_5-y_3 = (5\lambda D)/(d)-(3\lambda D)/(d)=(2\lambda D)/(d)`

where we used respectively m=3 and m=5. If we know solve for
`\lambda`, we find the wavelength:

`\lambda = ((y_5-y_3) d)/(2 D)=((5.5\cdot 10^(-3))(0.3\cdot 10^(-3)))/(2(1.5))=5.5 \cdot 10^(-7) m=550 nm`