A double-slit experiment is set up using red light (l = 712 nm). A first order b

Question

A double-slit experiment is set up using red light (l = 712 nm). A first order bright fringe is seen at a given location on a screen. What wave of visible light (between 380 nm and 750 nm) would produce a length dark fringe at the identical location on the screen? A new experiment is created with the screen at a distance of 2 m from the slits (with spacing 0.1 mm). What is the distance between the second order bright fringe of light with I = 697 nm and the third order bright fringe of light with I = 413 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

for bright fringe Y = n*lamda*D/d

for dark fringe Y = (n-0.5)*lamda*D/d

n1*lamda1 = (n2-0.5)*lamda2

n1 = 1 , lamda1 = 712 nm

lamda2 = 380 nm

1*712 = (n2-0.5)*380

n2 = 2.37 ..(1)

1*712 = (n2-0.5)*750

n2 = 1.449 ..(2)

from (1) and (2)

n2 = 2

(b) D =2 m , d = 0.1 mm

lamda1 = 697 nm , lamda2 = 413 nm

Y1 - Y2 = (2*697*10^-9*2/(0.1*0.001)) - ((3-0.5)*413*10^-9*2/(0.1*0.001))

= 0.00723 m

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