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If the entire apparatus of Exercise 35.9 (slits, screen, and space in between) is immersed in water, what then is the distance between the second and third dark lines?

Short Answer

Expert verified

The distance between the second and third dark lines of interference pattern when the setup is immersed inside water is0.625 mm.

Step by step solution

01

Given Data

Wavelength emitted: 500nm

Distance between slit and screen: 0.75m

Distance between slits: 0.45mm

Refractive index of water1.333

02

(a) Concept of Young’s double slit Experiment.

The redistribution of the intensity of light is when lights from two coherent sources having the same phase are superimposed on each other. This phenomenon is called interference of light.

The expression of location of dark bands here is,

ym=Rd ...(i)

Here,ymis the distance of the fringe,λis the wavelength,role="math" localid="1663937838953" mis the order of the fringe,Ris the distance between the screen and the slits, anddis the slit width.

03

(b) Determination of the separation between the two slits when the setup is immersed in water.

The wavelength of the light changes when the setup is immersed in water and this change is given as,

λ=λ0n

Here,λ0is the wavelength in air,λis the wavelength in water andnis the refractive index of water.

For water,n=1.333.

Distance between adjacent dark lines is given as,

y=ym+1-ym=Rλd=Rλ°dn

Substitute all the values,

role="math" localid="1663938770536" y=0.750m500×10-90.450×10-3m1.333=6.25×10-4=0.625mm

Thus, the distance between the fringes is 0.625 mm.

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