A Young’s double-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.48 mm. The interf

Question

A Young’s double-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.48 mm. The interference pattern on the screen 3.7 m away shows the first maximum 5.1 mm from the center of the pattern. What is the wavelength of the light in nm

in progress 0
Khánh Gia 6 months 2021-07-29T23:41:05+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-07-29T23:42:21+00:00

    Answer:

    Wavelength of light used will be equal to 661 nm

    Explanation:

    We have given distance between slits d = 0.48 mm = 0.48\times 10^{-3}m

    It is given interference pattern is 3.7 m away so D = 3.7 m

    First maximum from the center is 5.1 mm

    So y=5.1\times 10^{-3}m

    Distance of the first maximum from the center is equal to y=\frac{\lambda D}{d}

    5.1\times 10^{-3}=\frac{\lambda\times  3.7}{0.48\times 10^{-3}}=661\times 10^{-9}=661nm

    So wavelength of light used will be equal to 661 nm

Leave an answer

Browse

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