Answer 1

Distance of the screen from the slits, D = 1 m

Wavelength of light used, `lambda_1 = 600 nm`

Angular width of the fringe in air `theta_1 = 0.2^@`

Angular width of the fringe in water = `theta_2`

Refractive index of water, `mu = 4/3`

Refractive index is related to angular width as:

`mu = theta_1/theta/2`

`theta_2 = 3/4 theta_1`

`= 3/4 xx 0.2 = 0.15`

Therefore, the angular width of the fringe in water will reduce to 0.15°.