A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

Question

Deepak Patra · Answer

Sol. Given that, μ = 1.45, t = 0.02 mm = 0.02 * 10^–3 m and λ = 620 nm = 620 * 10^–9 m We know, when the transparent paper is pasted in one of the slits, the optical path changes by (μ – 1)t. Again, for shift of one fringe, the optical path should be changed by λ. So, no. of fringes crossing through the centre is given by, n = (μ – 1)t/λ = 0.45 * 0.02 * 10^-3/620 * 10^-9 = 14.5

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A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

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A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

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