Question

In Young's experiment, a pair of slits separated by a distance of 0.9 mm is placed at a distance of
1.0 m from the screen. The tenth bright fringe is at a distance of 6.0 mm from the centre of the
fringe pattern. What is the wavelength of light used? If the entire experimental set up is
immersed in a medium of refractive index μ, the tenth bright fringe is found to be at a distance
of 4.5 mm from the centre of the fringe pattern. Find μ.

Answer 1

Answer:

this is the answer of this question

Answer 2

Given,

Distance between pair of slits, d = 0.9 mm

distance of the slits from the screen, D = 1 m

Distance of 10th bright fringe from the centre of the fringe pattern, = 6 mm

Distance of 10th bright fringe from the centre of the fringe pattern(while in a medium of R.I. μ) = 4.5 mm

To Find,

i) the wavelength of light used, $\lambda$ =?

ii) refractive index of the medium, μ =?

Solution,

i) We know, in Young's experiment, the distance of the nth bright fringe from the centre of the fringe pattern,

$x_n = \frac{nD\lambda}{d}$

Therefore,

$x_1_0 = \frac{10D\lambda}{d} \\6 = \frac{10*1000*\lambda}{0.9}\\\lambda = \frac{6*0.9}{10000} \\\lambda = 54*10^{-3} mm\\\lambda = 54*10^{-6} m$

Hence, the wavelength of light used, $\lambda$ = 54*$10^-^6$ m

ii) If the apparatus of Young’s double-slit experiment is immersed in a liquid of refractive index (μ), then the wavelength of light decreases ‘μ’ times.

So in this case,

$x_{10} = \frac{10D\frac{\lambda}{\mu} }{d} \\6 = \frac{10*1000*\frac{54*10^{-3}}{\mu} }{0.9}\\\mu = \frac{540}{6*0.9} \\\mu = 100$

Hence, the refractive index of the medium, μ = 100.