Answer:
= 1220 nm
= 1.22 μm
Explanation:
given data:
wavelength [tex]\lambda = 610 nm = 610\times 10 ^{-9} m[/tex]
distance of screen from slits D = 3 m
1st order bright fringe is 4.84 mm
condition for 1 st bright is
[tex]d sin \theta =\lambda[/tex] ---( 1)
and[tex] tan \theta = \frac{y}{ D}[/tex]
[tex] \theta = tan^{-1}\frac{(y }{D})[/tex]
= 0.0924 degrees
plug theta value in equation 1 we get
[tex] d sin ( 0.0924) = 610 \times 10 ^{-9}[/tex]
[tex]d = 3.78\times 10^{-4} m[/tex]
condition for 1 st dark fringe
[tex] d sin \theta =\frac{λ'}{2}[/tex]
[tex]\lambda '= 2 d sin\theta[/tex]
= 2λ since from eq (1)
= 1220 nm
= 1.22 μm
