Question

Fraunhofer diffraction pattern is obtained by using a single slit and light of wavelength 6000A○.When the whole apparatus is immersed in a liquid,the angular width of the central macima reduces 20%.Calculate the Refractive index of the liquid.​

Answer 1

Answer:

i have not understood

Explanation:

Answer 2

The Refractive index of the liquid is 1.25

Given,

Light wavelength 6000A°

the angular width of the central macima reduces 20%.

To Find,

Find the Refractive index of the liquid.

Solution,

The angular width of central maxima is given by:

$θ_{w} = 2θ= \frac{2λ}{a}$

when the light of wavelength $λ_{1}$= 6000A° is used, $θ_{w_{1} } = \frac{2λ_{1} }{a}$ .....(1)

Angular Width of central maxima when light of wavelength is used,

$θ_{w_{2} } = \frac{2λ}{a}$ .....(2)

Now we are divide equation 2 and equation 1

$\frac{θ_{w_{2} } }{θ_{w_{1} } } = \frac{λ}{λ_{1} }$

=> $θ_{w_{2}$ = 0.8 $θ_{w_{1}$

=> $0.8= \frac{λ}{6000}$

=> λ = 0.8 × 6000 = 4960 A°

According to the snell's law,

$λ_{m} = \frac{λ}{u}$

Accordingly, angular fringe width when the apparatus is immersed in a liquid is given by

$θ_{w_{1} } = \frac{2λm}{w} = \frac{2λ}{ua}$

=> $θ_{w_{1} } = \frac{θ_{w} }{u}$

=> $0.8 = \frac{1}{u}$

=> u= 1.25

Hence, the Refractive index of the liquid is 1.25

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For more information about Fraunhofer diffraction, you can refer to the following questions

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