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Light is incident normally on two narrow parallel slits a distance of 1.00 mm apart. A screen is placed a distance of 1.2 m from the slits. The distance on the screen between the central maximum and the centre of the n=4 bright spot is measured to be 3.1 mm. a Determine the wavelength of light. b This experiment is repeated in water (of refractive index 1.33). Suggest how the distance of 3.1 mm would change, if at all.

Respuesta :

Answer:

Explanation:

distance between slits d = 1 x 10⁻³ m

Screen distance D = 1.2 m

Wave length of light   = λ

Distance of n th bright fringe fro centre

= n λ D / d where n is order of bright fringe . Here n = 4

Given

3.1 x 10⁻³ = (4 x λ x 1.2) / 1 x 10⁻³

λ = 3.1 x 10⁻⁶ / 4.8

= .6458 x 10⁻⁶

6458 x 10⁻¹⁰m

λ= 6458 A.

The distance will reduce 1.33 times

New distance = 3.1 /1.33

= 2.33 mm.

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