A double-slit experiment is set up using red light ( = 736 nm). A first order br

Question

A double-slit experiment is set up using red light ( = 736 nm). A first order bright fringe is seen at a given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?

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A new experiment is created with the screen at a distance of 1.9 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with = 684 nm and the third order bright fringe of light with = 404 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)

Explanation / Answer

(a)for bright fringe ym = mD/d D= distance between slits and screen and d= distance between the slits, lambda = wavelength

for first order bright fringe m = 1,

          y1 = (1)D/d

          y1 = (736 nm)D/d   

for dark fringe,we have

        ym = (m + 1/2)D/d   

Now to get the dark fringes at the same location we should have;

     (736nm)D/d = (m + 1/2)D/d    

put m = 1

      (1 + 1/2) = (736 nm)

= 490.7 nm

(b) first wavelength 1 =684 nm = 684*10-9 second wavelength 2 = 404 nm = 404*10-9 m

distance between slits and screen D = 1.9 m

slits separation d = 0.0.08*10-3 m

for second order bright (m = 2) for first wave length ,

                 y2 = 21D/d

for third order bright (m = 3) for first wave length ,

                 y3 = 32D/d

y2 - y3 = (21 - 32)D/d   .

     y2 - y3 = (2*684*10-9 - 3*404*10-9)(1.9) / (0.08*10-3)

                = 3.705*10-3 m

                = 3.705 mm

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