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g When a laser light of unknown wavelength is passed through two narrow slits separated by 150 μm, an interference pattern is observed on a screen 1.5 m away. The distance between the central spot and first order constructive interference is found to be 5 mm. a) Find the wavelength of the laser light, in nanometers. (b) At what angle, in degrees, from the central spot is the second-order constructive interference formed?

Answer :

Answer:

(a) 500 nm

(b) [tex]\theta = 0.382^{\circ}[/tex]

Solution:

Slit separation, d = [tex]150\mu m = 150\times 10^{- 6}\ m[/tex]

Distance from the screen, x = 1.5 m

Distance between central spot and 1st order constructive interference, [tex]y_{1} = 5 mm = 5\times 10^{- 3}\ m[/tex]

Now,

(a) To calculate the wavelength of laser light:

From Double-slit experiment, we know that the distance between the central fringe and the nth bright fringe is given by:

[tex]y_{n} = \frac{n\lambda x}{d}[/tex]

where

n = 1

[tex]\lambda[/tex] = wavelength of the light used

Thus

[tex]y_{1} = \frac{1.\lambda x}{d}[/tex]

[tex]5\times 10^{- 3} = \frac{\lambda \times 1.5}{150\times 10^{- 6}}[/tex]

[tex]\lambda = 5\times 10^{- 7}\ m = 500\ nm[/tex]

(b) To calculate the angle for the formation of the constructive interference of second order:

Here

n = 2

Thus

[tex]y_{2} = \frac{2\lambda x}{d}[/tex]

[tex]tan\theta = \frac{\frac{2\lambda x}{d}}{x} = \frac{2\lambda }{d}[/tex]

[tex]tan\theta = \frac{2\times 500\times 10^{- 9}}{150\times 10^{- 6}}\ m[/tex]

[tex]\theta = tan^{- 1} (6.67\times 10^{- 3})[/tex]

[tex]\theta = 0.382^{\circ}[/tex]

Lanuel

The wavelength of the laser light in nanometers is equal to 5 nanometer.

Given the following data:

Distance between slits = 150 um to m= [tex]150 \times 10^{-6}\;m[/tex]

Distance from screen = 1.5 m.

Distance, [tex]y_1[/tex] = 5 mm to m = 0.005 m.

How to calculate the wavelength.

In order to calculate the wavelength of the laser light, we would apply the double-slit interference experiment.

Mathematically, the double-slit interference experiment is given by this formula:

[tex]y_n=\frac{\lambda mx}{d}[/tex]

Where:

  • d is the distance between slits.
  • m is the order of fringe.
  • [tex]\lambda[/tex] is the wavelength.

Substituting the given parameters into the formula, we have;

[tex]0.005=\frac{\lambda \times 1 \times 1.5}{150 \times 10^{-6}} \\\\0.005=\frac{1.5 \lambda}{150 \times 10^{-6}}\\\\1.5 \lambda = 150 \times 10^{-6} \times 0.005\\\\1.5 \lambda =7.5 \times 10^{-7}\\\\ \lambda =\frac{7.5 \times 10^{-7}}{1.5} \\\\ \lambda =5 \times 10^{-7}[/tex]

Note: 1 nanometer = [tex]1\times 10^{-7}[/tex] meter.

Wavelength = 5 nanometer.

How to calculate the angle.

Mathematically, the angle from the central spot at which the second-order constructive interference is formed is given by:

[tex]Tan \theta = \frac{2\lambda}{d} \\\\\theta = tan^{-1}(\frac{2\lambda}{d}) \\\\\theta = tan^{-1}(\frac{2 \times 5\times 10^{-7}}{150 \times 10^{-6}})\\\\\theta = tan^{-1}(6.67 \times 10^{-3})[/tex]

Angle = 0.382°.

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