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Exercises · 10.7

Q.In a double-slit experiment the angular width of a fringe is found to be 0.2∘0.2^\circ on a screen placed 1 m1\ \text{m} away. The wavelength of light used is 600 nm600\ \text{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 43\frac{4}{3}.

Puducherry CbseNCERTSubjective· 2mImportance★★★★★
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The angular fringe width is θ=λ/d\theta = \lambda/d; immersing the setup in water reduces the wavelength to λ/μ\lambda/\mu, so the new angular width is 0.2∘×34=0.15∘0.2^\circ \times \frac{3}{4} = 0.15^\circ.

Step 1: Angular fringe width formula

In Young's double-slit experiment, the linear fringe width on screen is β=λDd\beta = \dfrac{\lambda D}{d}, so the angular fringe width (independent of the screen distance DD) is

θ=βD=λd\theta = \frac{\beta}{D} = \frac{\lambda}{d}

Step 2: Wavelength change in water

The frequency of light does not change when it enters a new medium, but its speed and hence wavelength do:

λwater=λairμwater=600 nm4/3=450 nm\lambda_{water} = \frac{\lambda_{air}}{\mu_{water}} = \frac{600\ \text{nm}}{4/3} = 450\ \text{nm}

Step 3: New angular fringe width

Since dd (the slit separation) is unchanged, and θ∝λ\theta \propto \lambda: …

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