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A slit of width ‘d’ is illuminated by light of wavelength 5000 Å. For what value of ‘d’ will the first maximum fall at an angle of diffraction of 30˚. |
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Answer» Here, λ = 5000Å = 5× 10−7m d = ?; n = 1 ; θ = 30˚ For maxima of diffraction, d sin θ = (2n + 1)\(\frac{λ}{2}\) or, d = \(\frac{(2n+1)\lambda}{2sin\theta}\) = \(\frac{3\times 5\times 10^{-7}}{2sin\,30°}\) = 1.5× 10−6 m. |
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