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10 November, 21:30

Using the equation for the distance between fringes, Δy = xλ d, complete the following. (a) Calculate the distance (in cm) between fringes for 694 nm light falling on double slits separated by 0.0850 mm, located 4.00 m from a screen. cm (b) What would be the distance between fringes (in cm) if the entire apparatus were submersed in water, whose index of refraction is 1.333? cm

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  1. 11 November, 00:16
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    Distance between fringe or fringe width = xλ / d

    where x is location of screen and d is slit separation

    Given x = 4 m

    λ = 694 nm

    d =.085 x 10⁻³ m

    distance between fringes

    = 4 x 694 x 10⁻⁹ /.085 x 10⁻³

    = 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶

    = 32.66 x 10⁻³ m

    = 32.66 mm.

    3.267 cm

    b)

    when submerged in water, wavelength in water becomes as follows

    wavelength in water = wave length / refractive index

    = 694 / 1.333 nm

    = 520.63 nm

    new distance between fringes

    3.267 / 1.333

    = 2.45 cm.
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