coherent monochromatic light has normal incidence on a barrier with two parallel

Question

coherent monochromatic light has normal incidence on a barrier with two parallel slits separated by a distance of 2.00mm. the diffraction pattern produced by the slits appears on a screen 5.00m away from the barrier. the screen is parallel to the plane of the barrier. The distance between the first diffraction maximum on one side of the central maximum and the second diffraction minimum on the other side is 3.40mm. (make a sketch). what is the wavelength of the light?

Explanation / Answer

Maxima are at angle theta given by, sin(theta) =m*wavelength/d and on a screen, a distance L away, the maxima are at a distance y away from the centre where y = L*tan(theta) If the angle theta is small sin(theta) =tan(theta) so y/L = m*wavelength/d => wavelength = yd/mL Now, for m = 1, we know that y= 3.40 mm. Therefore, wavelength = 3.4*10^-3*2*10^-3/1*5 = 1.36*10^-6 m = 1360 nm

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