Consider a plane wave of monochromatic green light, lambda = 500 nm, that is inc

Question

Consider a plane wave of monochromatic green light, lambda = 500 nm, that is incident normally on two identical narrow slits (the widths of the individual slits are much less than the wavelength of the light). The slits are separated by a distance of d = 30 mu m. An interference pattern is observed on a screen located a distance L away from the slits. On the screen, the location nearest the central maximum where the intensity is zero (i.e., the first dark fringe) is found to be 1.5 cm from this central point. Let this particular position on the screen be referred to as P1. Calculate the distance L to the screen. Show all work. Calculate the distance between the first and second dark bands in the interference pattern. Show all work. In each of the parts below, one change has been made to the problem above (in each case, all parameters not explicitly mentioned have the value or characteristic stated above changes are not cumulative). For each case, explain briefly whether the light intensity at location P1 would remain zero or not If not, will P1 become a location of maximum constructive interference (a bright fringe)? In each case, explain your reasoning. The wavelength is doubled so that lambda = 1000 nm. The slits are moved closer together, so that they are at a distance of d = 15 mu m.

Explanation / Answer

a)

Lambda*D/2d = 1.5 * 10^-2

500 * 10^-9 * D / 2 * 30*10^-6 = 1.5 * 10^-2

D = 1.8 m

b)distance bet dark fringes = D*lambda/d = 1.8 * 500*10^-9 / 30*10^-6 = 0.03 m = 3 cm

c)

i)

bright fringe = D*lambda/d = 3 cm

so P1 will not be a right fringe , it will be a dark fringe , intensity will be 0

ii)

bright fringe = D*lambda/d = 3 cm

so P1 will not be a right fringe , it will be a dark fringe , intensity will be 0

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