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

Answer:

550 nm

Explanation:

The formula for the double-slit diffraction experiment is:

$y=\frac{m\lambda D}{d}$

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:

$y_5-y_3 = \frac{5\lambda D}{d}-\frac{3\lambda D}{d}=\frac{2\lambda D}{d}$

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

$\lambda = \frac{(y_5-y_3) d}{2 D}=\frac{(5.5\cdot 10^{-3})(0.3\cdot 10^{-3})}{2(1.5)}=5.5 \cdot 10^{-7} m=550 nm$