Question

Light of wavelength 5.0 x 10–7 meter in air is incident normally (perpendicularly) on a double slit. The distance between the slits is 4.0 x 10–4 meter, and the width of each slit is negligible. Bright and dark fringes are observed on a screen 2.0 meters away from the slits.

a. Calculate the distance between two adjacent bright fringes on the screen.

The entire double-slit apparatus, including the slits and the screen, is submerged in water, which has an index of refraction 1.3.

b. Determine each of the following for this light in water.

i. The wavelength

ii. The frequency

c. State whether the distance between the fringes on the screen increases, decreases, or *** the same. Justify your answer.

Answer 1

(a) Simple application of the formula, m λ = d x / L … (1)(5x10–7) = (4x10–4)(x) / (2) … x = 2.5x10–3 m

(b) i) n1 λ1 = n2 λ2 … (1)(5x10–7) = (1.3) λwater … λwater = 3.85 x10–7m

ii) frequency does not change when changing mediums, same as air.

c = fair / λair … 3x108 = fair / 5x10–7 … fair = 6x1014 Hz = fwater

(c) Based on mλ = d x / L, since the λ is less, the x is less also which means the fringe spacing has decreased.