Question

(1A) Light of wavelength 534.0 nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of the m = 82.0 bright fringe, how much farther is it to the more distant slit than to the nearer slit?

(1B) Light from a sodium lamp of wavelength 446.0 nm illuminates two narrow slits. The fringe spacing on a screen 137.8 cm behind the slits is 6.21 mm. What is the spacing between the two slits?

(1C) Light is incident normal to the surface of a 1.07 cm layer of water that lies on top of a flat Lucite plate with a thickness of 0.560 cm. How much more time is required for light to pass through this double layer than is required to traverse the same distance in air? (nLucite = 1.61, nwater = 1.333)

Answer 1

1A) The wavelength = 534.0 nm , The order of bright fringe m = 82.0

The path difference for mth bright fringe is given by

putting the value we get

path diff = 534.0 X 82 = 43788 nm = 4.38 X 10^-5 m

1B) The wavelength = 446.0 nm.

The distance between the screen and slit D is 137 cm

The fringe width is = 6.21 mm

the fringe width is given by

where the slit separation is d

putting the values in the above eqn we get

d= 9.84 X 10-5 m or 0.0984 mm

1C) The refractive index of water

the speed of light in water is

where c is the speed of light in vaccuam c = 3 X 10 ⁸ m/s

putting the values we have v = 2.25 X 10 ⁸ m/s  

The water is 1.07 cm thick .

The time taken to pass through water is T1

putting the values we get

T1 = 4.75 x 10 ^-11 s

The refractive index of nLucite  

the speed of light in water is

where c is the speed of light in vaccuam c = 3 X 10 ⁸ m/s

putting the values we have v = 1.875 X 10 ⁸ m/s  

The nLucite is 0.560 cm thick .

The time taken to pass through water is T2

putting the values we get

T2 = 2.7 x 10 ^-11 s

hence the total time required to pass the double layer is

T = T1+ T2 = 7.45 X 10 ^-11 s