Question

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.350mm . This results in an interference pattern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits.

Part A

What is the wavelength ?2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d = 0.350mm ?

Express your answer in millimeters.

?2 =

Answer 1

given
wavelength of first laser ?1 = d/8
slits seperation d1 = 0.35mm
distance between the slits to the screen is L
the distance of minimum from one end of thecentral maximum is
Y = (2m+1)?L / 2d ------------(1)
so for fourth minima m = 4
so Y4 =9?1L/2d1

the distance of next maximum from one end ofthe central maximum is
Y = m?L / d
so for second maximum m =2
so Y2 =2?2L/d2 ---------------(2)
so from (1) & (2)
9?1L/2d1 =2?2L/d2
or 9?1/2d1 =2?2/d2
or 9d 1/ 8*2d1 =2?2/d2
as the slit seperation is same for both thelasers ( i.e d1= d2 = d =0.3500mm),so we have
?2 = 9d2/32d
or ?2 = 9d/32

?2=9(0.3500)/32=9.84 x 10^-2