Question

15. A double-slit arrangement produces
interference fringes for sodium light
(1 = 589 nm) that are 0.20° apart. What
is the angular fringe separation if the
entire arrangement is immersed in water
(n = 1.33)?
[Ans: 0.159]​

Answer 1

Given that,

Wavelength = 589 nm

Width = 0.20°

We know that,

The angular fringe separation in air is

$\beta_{a}=\dfrac{\lambda_{a}D}{2d}$....(I)

The angular fringe separation in water is

$\beta_{w}=\dfrac{\lambda_{w}D}{2d}$......(II)

We need to calculate the angular fringe separation

Using formula of refractive index

$\mu=\dfrac{\text{speed of light in air}}{\text{speed of light in medium}}$

$\mu=\dfrac{v_{a}}{v_{w}}$

$\mu=\dfrac{\lambda_{a}f}{\lambda_{w}f}$

$\mu=\dfrac{\lambda_{a}}{\lambda_{w}}$

$\dfrac{4}{3}=\dfrac{\lambda_{a}}{\lambda_{w}}$

$\lambda_{w}=\dfrac{3}{4}\lambda_{a}$

Put the value of $\lambda_{w}$ in the equation

$\beta_{w}=\dfrac{\dfrac{3}{4}\lambda_{a}D}{2d}$

$\beta_{w}=\dfrac{3}{4}\beta_{a}$

$\beta_{w}=\dfrac{3}{4}\times0.20$

$\beta_{w}=0.15$

Hence, The angular fringe separation is 0.15

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Topic : double slit interference

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