Question

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light.

Answer 1

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

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

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$