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14 January, 20:26

A thin slit illuminated by light of frequency f produces its first dark band at ± 38.2 ∘ in air. When the entire apparatus (slit, screen and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at ± 17.3 ∘.

Find the refractive index of the liquid.

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  1. 14 January, 23:59
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    the refractive index of the liquid is 2.08

    Explanation:

    The diffraction formula is:

    d sin (θ) = mλ

    where

    d is the diffraction m = 1, 2 λ is the wavelength of the light

    Thus,

    d sin (θ₁) = m λ₁ (1) and

    d sin (θ₂) = m λ₂ (2)

    Dividing equation (1) by equation (2) you get

    sin (θ₁) / sin (θ₂) = λ₁ / λ₂ (3)

    λ can be expressed as

    λ = v / f

    where

    v is the speed of light f is the frequency of light

    Therefore,

    λ₁ = v₁ / f

    λ₂ = v₂ / f

    Thus,

    λ₁ / λ₂ = v₁ / v₂

    Therefore, substituting the above expression for λ₁ / λ₂ into equation (3), we get

    sin (θ₁) / sin (θ₂) = v₁ / v₂

    where

    v₁ is the speed of light in air

    The expression for the refractive index is

    n = c / v, which can also be expressed as

    n = v₁ / v₂

    Therefore,

    n = sin (θ₁) / sin (θ₂)

    n = sin (38.2°) / sin (17.3°)

    n = 2.08

    Therefore, the refractive index of the liquid is 2.08
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