Question

To test some of the principles of physics a team of scuba divers sets up a double slit experiment under water in a swimming pool. Red light is emitted by a laser (kept out of the water), and has a wavelength of 632.8 nm in air. The laser beam is directed by mirrors to a double slit located near the bottom of the pool. Just before the light strikes the double slit it is traveling horizontally about a 1⁄2 meter above the pool bottom. The light passes through the double slit and forms an interference pattern on the side wall of the pool, 5.0 m from the slits, as measured in the direction of the beam just before it reaches the slits. The slits are oriented vertically, and are 0.60 mm apart. a. What is the wavelength of the laser light when traveling through the water? (index of refraction = 1.33) b. Calculate the separation between central maximum and the first order maximum of the interference pattern, as measured on the side wall of the pool. c. What happens to the interference pattern if the pool is drained? (A numerical answer is not necessary.) Justify your answer.

Answer 1

given
wavelength of light in air, lamda = 632.8 nm
slit separation, d = 0.6 mm = 0.6*10^-3 m
R = 5.0 m

a) wavelength of light in water, lamda' = lamda/n

= 632.8/1.33 nm

= 475.8 nm <<<<<<<--------------Answer

b) the separation between central maximum and the first order maximum of the interference pattern = lamda'*R/d

= 475.8*10^-9*5/(0.6*10^-3)

= 3.96*10^-3 m (or) 3.96 mm <<<<<<<--------------Answer

c) The interference pattern size will be increases.

the separation between central maximum and the first order maximum of the interference pattern = lamda*R/d

= 632.8*10^-9*5/(0.6*10^-3)

= 5.27*10^-3 m (or) 5.27 mm <<<<<<<--------------Answer