Question

A beam of green light is diffracted by a slit with a width of 0.500 mm. The diffraction pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 3.10 mm. Calculate the wavelength of the laser light. (answer in nano meters)

Answer 1

first you need to convert unit into metres and then continue with calculation

Answer 2

Answer: 376.2 nm

Step-by-step explanation:

The equation for single slit diffraction is
`wsin(\theta)=m\lambda` (w is width of slit,
`\theta` is the angle of the fringe of concern to the slit, m is a number based on the type of fringe it is, and
`\lambda` is the wavelength of the light)

The distance between the slit and the wall is given, and the distance between the fringes of destructive interference that are next to the central bright spot is also given.

Thus, trigonometry can be used to find the angle of one of the fringes of destructive interference of concern to the slit.

The distance between the central bright fringe and a fringe of destructive interference next to the central bright fringe is
`(3.10*10^(-3))/(2)=1.55*10^(-3)` meters.

Thus,
`tan(\theta)=(1.55*10^(-3))/(2.06)`, which then means that
`\theta=tan^(-1)((1.55*10^(-3))/(2.06))=0.0431` degrees.

We know the width of the slit.

Since the fringe of destructive interference is next to the central bright fringe, the value of m is 1. m = 1,2,3,... for destructive interference fringes and m=1/2,3/2,5/2,... for constructive interference fringes. m increases by 1 for every successive constructive or destructive fringe that gets further from the central bright fringe.

Plugging into the equation gives
`5*10^(-4)*sin(0.0431)=1*\lambda`.

Rearranging the equation gives
`\lambda=5*10^(-4)*sin(0.0431)=3.762*10^(-7)` meters.

Converting to nanometers gives
`\lambda=3.762*10^(-7)*10^9=376.2` nanometers.