Question

In a single-slit diffraction experiment, the width of the slit is 1.90 μm. If a beam of light of wavelength 610 nm forms a diffraction pattern, what is the angle associated with the second dark fringe above the central bright fringe?

a) 39.9 degrees
b) 11.4 degrees
c) 18.7 degrees
d) 12.2 degrees

Answer 1

The the angle associated with the second dark fringe is 39.9°

Option (a) is correct.

Step-by-step explanation:

Given:

Wavelength of light
`\lambda = 610 * 10^(-9)` m

Width of the slit
`a = 1.90 * 10^(-6)` m

Order of diffraction
`m = 2`

From the formula of single slit diffraction,

`a\sin \theta = m \lambda`

Where
`m =` order of diffraction,
`\theta =` angle associated with second dark fringe.

`\sin \theta = (\lambda )/(a)`

`\sin \theta = (2 * 610 * 10^(-9) )/(1.90 * 10^(-6) )`

`\sin \theta = 0.6421`

`\theta = \sin ^(-1) (0.6421)`

`\theta =` 39.9°

Therefore, the the angle associated with the second dark fringe is 39.9°