a source of light having wavelength 5500a.u is falls normally on slit a width of 22*1/100000cm.calculate the angular seperation of the first two minima on either side of central maxima.
Anonymous:
What's the separation between screen and source?
Oh! Sorry it isn't needed. My mistake.
Answers
Answered by
2
You need the fringe length ß = λD/d.
And angular separation is θ = ß/D = λ/d = (55/22)/(10) = 0.25 radians
Answered by
1
You should write the wavelength as °A (Angstroms) and not as a.u. the unit a.u. is used in astronomy and it is very large.
In the single slit experiment of Young, we have the angle Ф of the minimum intensities on the screen with respect to the perpendicular bisector of the slit as:
sin Ф ≈ Ф ≈ n λ / a , n = +- 1, +- 2, +- 3,....
where λ = wavelength of the light wave of single frequency
a = single slit width
hence the angular separation of the first two minima = 2λ/a - λ/a = λ/a
= 5500 * 10⁻⁸ cm / 22 * 10⁻⁵ cm
= 0.25 radians = 14.32 degrees
In the single slit experiment of Young, we have the angle Ф of the minimum intensities on the screen with respect to the perpendicular bisector of the slit as:
sin Ф ≈ Ф ≈ n λ / a , n = +- 1, +- 2, +- 3,....
where λ = wavelength of the light wave of single frequency
a = single slit width
hence the angular separation of the first two minima = 2λ/a - λ/a = λ/a
= 5500 * 10⁻⁸ cm / 22 * 10⁻⁵ cm
= 0.25 radians = 14.32 degrees
click on thank you.
Similar questions