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

Question

A double-slit experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 1.6 mm apart on the viewing screen.What will the fringe spacing be if the light is changed to a wavelength of 360nm?

in progress 0
Hải Đăng 6 months 2021-08-05T21:49:58+00:00 1 Answers 6 views 0

Answers ( )

    0
    2021-08-05T21:51:34+00:00

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )