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6 June, 19:09

Light passes through a single slit of width 8.32x10^-5 m. The first (m=1) diffraction minimum occurs at an angle of 0.347. What is the wavelength of the light in nanometers?

(Hint: The answer will be between 400 and 700 nm.)

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  1. 6 June, 22:34
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    theta = m wavelength / silt width

    gives angular position of minimas where m = 1,2,3 ...

    As

    0.347° = pi/180 * 0.347 rad = 0.006 rad

    using above formula

    0.006 = 1 * wavelength / 8.32 * 10^-5 m

    wavelength = 0.006 * 8.32 * 10^ - 5 m

    = 0.04992 * 10^ - 9 m

    = 499 * 10^ - 9m

    =499 nm

    which is almost 500 nm
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