標題:
Young's double slit experiment
發問:
At Young's double slit experiment is carried out in a liquid whose refractive index is 1.2 . The distance between two successive bright fringes on the screen is 0.5mm. If the whole set-up is placed in another liquid whose refractive index is 1.5, what will be the distance between two successive bright fringes?
最佳解答:
The formula for fringe separation y in Young's double slit experiment is, y = 入D/(n.a) where 入 is the wavelength of light used. D is the distance of the slits from the "screen" n is the refractive index of the medium a is the slit separation Hence, 0.5 = 入D/1.2a and y = 入D/1.5a (where y is the new fringe separation) Dividing, y/0.5 = 1.2/1.5 y = (1.2/1.5) x 0.5 m = 0.4 m 2013-08-04 19:14:08 補充： sorry for a typo... the unit for fringe separation should be mm. Hence, the answer is 0.4 mm
Young's double slit experiment
發問:
At Young's double slit experiment is carried out in a liquid whose refractive index is 1.2 . The distance between two successive bright fringes on the screen is 0.5mm. If the whole set-up is placed in another liquid whose refractive index is 1.5, what will be the distance between two successive bright fringes?
最佳解答:
The formula for fringe separation y in Young's double slit experiment is, y = 入D/(n.a) where 入 is the wavelength of light used. D is the distance of the slits from the "screen" n is the refractive index of the medium a is the slit separation Hence, 0.5 = 入D/1.2a and y = 入D/1.5a (where y is the new fringe separation) Dividing, y/0.5 = 1.2/1.5 y = (1.2/1.5) x 0.5 m = 0.4 m 2013-08-04 19:14:08 補充： sorry for a typo... the unit for fringe separation should be mm. Hence, the answer is 0.4 mm
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