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stealth61 [152]
4 years ago
5

A double-slit experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 1.6 mm apart o

n the viewing screen.What will the fringe spacing be if the light is changed to a wavelength of 360nm?
Physics
1 answer:
yuradex [85]4 years ago
7 0

Answer:

1.44*10^-3m

Explanation:

Given that distance BTW two bright fringes is

DetaY = lambda* L/d

So for second wavelength

Deta Y2= Lambda 2* L/d

=lambda 2 x deta y1/ lambda1

So substituting

= 360 x 10^-9 x (1.6*10^-3/640*10^-9)

1.44*10^ -3m

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