In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

n^{th} order Angular width of fringe in air,

sin(θ_{n}) = nλ/d … (1)

Where sin(θ) is angular fringe width

n is order of fringe

λ is wavelength of light used

d is slit width

When entire apparatus is taken into water, d and D remain same

∴ Angular fringe width for nth order in water,

sin(θ_{n}^{w}) = nλ^{w}/d … (2)

where λ^{w} is wavelength of the light in water

Taking ratio of equations (1) & (2)

sin(θ_{n})/sin(θ_{n}^{w}) = λ/ λ^{w}

Refractive index of water is 4/3

i.e., 4/3 = c/v

Where c is velocity of light in air (vacuum) and v is velocity in water

c/v = ν λ /ν λ^{w}

Since frequency does not vary with medium, ν = ν^{w}

Thus 4/3 = λ/λ^{w}

Thus we have,

sin(θ)/sin(θ^{w}) = 4/3

sin(θ^{w}) = sin(θ) × 3/4

= 0.2 × 3/4 = 0.15°

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