A Young’s double slit apparatus has the slit separation of 0.15mm and the distan

Question

A Young’s double slit apparatus has the slit separation of
0.15mm and the distance between the slits and the viewing
screen is 150 cm.
A) If monochromatic light of wavelength ? = 550x10-9m passes
through the slits, calculate the distance between the central
(bright) fringe and the adjacent bright fringe.
B) Now imagine that you are using X-rays, ?=1x10–9m instead of
visible light in this experiment. What would the slit separation
need to be in order to have the distance between these first two
bright fringes be 1cm? '

detailed equations with work please, ive had this answered before but im still confused slightly.

Explanation / Answer

for bright fringes

dsin =m wher d is the distance between the slits

hence A) .00015 sin= 550*10^(-9) since m=1 for first bright fringe

sin=3.666 *10^(-3)

also sin=y/D where y is the distance between the central and first bright fringe

and D is the distance between the slits and the screen

implies y=1.5*3.666*10^(-3) =5.5 *10^(-3) m

B)y= 1 cm=.01 m

hence sin= .01/1.5=6.666*10^(-3)

dsin= implies d=10^(-9)/6.666*10^(-3) =1.5 *10^(-7) m

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