A low-pressure sodium vapor lamp whose wavelength is 5.89 x 10^−7 m passes through double-slits that are 2.0 x 10^−4 m apart and produces an

Question

A low-pressure sodium vapor lamp whose wavelength is 5.89 x 10^−7 m passes through double-slits that are 2.0 x 10^−4 m apart and produces an interference pattern whose fringes are 3.2 x 10^−3 m apart on the screen. What is the distance to the screen?

3 meters

2.7 meters

1.6 meters

1.1 meters

its literally 1.1 meters I got it right

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Adela 2 months 2021-08-03T12:57:16+00:00 2 Answers 6 views 0

Answers ( )

    0
    2021-08-03T12:58:37+00:00

    Answer:

    Distance of the screen is approx 1.1 m

    Explanation:

    As we know that fringe width on Young’s double slit experiment is given as

    \beta = \frac{\lambda L}{d}

    here we know that

    \beta = 3.2 \times 10^{-3}

    \lambda = 5.89 \times 10^{-7} m

    d = 2.0 \times 10^{-4} m

    now we have

    3.2 \times 10^{-3} = \frac{(5.89 \times 10^{-7})L}{2.0 \times 10^{-4}}

    L = 1.1 m

    0
    2021-08-03T12:58:47+00:00

    Answer:

    it is 1.1 meters

    Explanation:

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