Light of wavelength 420nm goes through a single narrow slit of width a. The angl

Question

Light of wavelength 420nm goes through a single narrow slit of width a. The angle between the first minimum on the left and the first minimum on the right is 2 degrees. What is d (in nm)? A double slit diffraction pattern has an intensity envelop around the central maximum. And inside it there are n fringes. If the wavelength is 420nm and d=0.200 nm and a = 45 micrometers, then what is n? A double slit diffraction pattern has an intensity envelope around the central maximum. And inside it there are n fringes. If the wavelength is 420nm and d=0.200 nm and a = 45 micrometers, then what is the ratio of the intensity of the central fringe compared to the third fringe? (This should be bigger than one.) Young's Experiment is performed with wavelength 420nm and the slits are 2mm apart. How far away should you place a screen so that the closest fringes are 2mm apart? You wish to coat some glass (n=1.6) with a thin film of transparent material (n=1.33) so that light of wavelength 420nm (in vacuum) does not reflect. What minimum thickness (in nm) do you need? For thin films when you go from a lower index to a higher than the reflected wave picks up a phase shift of 0.5 wavelength. So going from air to the thin film the reflected wave gets 0.5 wavelength. The wave going through to film travels the thickness, then reflects (and picks up a 0.5 wavelength of phase) then travels the thickness and then transmits through (no change in phase). And to cancel the reflection you want a half wavelength total of phase difference. So (0.5 wavelength +2dn)-0.5 wavelength= odd multiple of wavelength/2. In Young's experiment the maxima are at d sin theta=m lambda where d is the distance between the thin slits and theta is the angle from the center and m is 0,1, 2 ... etcetera. In Young's experiment the minima are at d sin theta=(m+0.5) lambda where d is the distance between the thin slits and theta is the angle from the center and m is 0,1, 2 ... etcetera For single slit diffraction of width a has minima at angles that satisfy a sin theta=m lambda for m=l,2,3 ... For double slit diffraction (two slits each of width a, a distance d apart) when the intensity at the center is Im (intensity in the middle), then for an angle theta from the center you can compute alpha = pia/lambda sin theta and beta = pi beta/gamma sin theta then the intensity is given by l(thtea)=(I_m)((cos beta)^2) ((sin alpha/alpha)^2) The one with alpha is the intensity envelope, and the one with beta puts all the many fringes within each envelop that the alpha part makes.

Explanation / Answer

1) Here, d * sin(theta) = (n + 0.5) * lambda

=>   d * sin2 = (1 + 0.5)* 420

=> d = 18051.83 nm

2)    n = (0.200 * 45 *1000 )/(420)

            =   21

3)    ratio of intensity = 9/1

4) Far away = ( 2 * 2 * 10-6)/(420 * 10-9)

                     = 9.523 m

5)    minimum thickness = (420)/(2 * 1.33)

                                        = 157.89 nm

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