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23 December, 15:05

Helium atoms emit light at several wavelengths. Light from a helium lamp illuminates a diffraction grating and is observed on a screen 50.00 cm behind the grating. The emission at wavelength 501.5 nm creates a first-order bright fringe 21.90cm from the central maximum. What is the wavelength of the bright fringe that is 31.60cm from the central maximum?

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  1. 23 December, 15:43
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    wavelength = 667.94 nm

    Explanation:

    Given data

    screen distance L = 50.00 cm = 0.50 m

    wavelength = 501.5 nm

    bright fringe d = 21.90 cm = 0.2190 m

    bright fringe D = 31.60 cm = 0.3160 m

    to find out

    wavelength

    solution

    we will apply here triangle law

    tan (θ) = d / L

    (θ) = tan^-1 (0.219 / 0.50)

    (θ) = 23.65 degree

    so

    d sin (θ) = 1 (wavelength)

    d = 501.5 / sin (23.65)

    d = 1250.1 nm

    and

    for another angle

    tan (θ) = d / L

    (θ) = tan^-1 (0.316 / 0.50)

    (θ) = 32.3 degree

    so

    d sin (θ) = 1 (wavelength)

    wavelength = 1250 sin (32.3)

    wavelength = 667.94 nm
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