Question

In the double slit experiment using light of wavelength 600 nm ,the angular width of the fringe formed on the

Answer 1

Correct question:-

In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water?

Answer:-

We know that:-

→ Distance, D = 1m

→ Wavelength, $\lambda_{1}$ = 600 nm

Angular width of the fringe in air $\theta_{1}$ = 0.2°.

Angular width of the fringe in water = $\theta_{2}$

Refractive index of water, $\mu = \frac{4}{3}$

$\mu = \frac{\theta_1}{\theta_2}$is the relation between the refractive index and the angular width.

$\theta_2 = \frac{3}{4}\theta_1 \: \frac{3}{4} \times 0.2 = 0.15$

Therefore, 0.15° is the reduction in the angular width of the fringe in water.

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\blue{Correct\:Question:}}}}}}$

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$\hookrightarrow$ In a double slit experiment using light of wavelength 600nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?

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$\Large{\pink{\underline{\underline{\tt{\green{Solution:}}}}}}$

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$\textsf{Wavelength of light used,}$ $\sf \lambda\:=\:6000nm\:=\:600\times 10^{-9}m$

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$\textsf{Angular width of the fringe,}$

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$\sf \theta\:=\:0.1^{\circ}\times \dfrac{\pi}{180}\:=\: \dfrac{3.14}{1800}rad$

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Angular width of a fringe is related to slit spacing(d) as:

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$\sf \theta\:=\: \dfrac{\lambda}{d}$

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$\sf d\:=\: \dfrac{\lambda}{\theta}$

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$\sf d\:=\: \dfrac{600\times 10^{-9}}{\dfrac{3.14}{1800}}$

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$\sf d\:=\: \dfrac{600\times 10^{-19}\times 1800}{3.14}$

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$\sf d\:=\:3.44\times 10^{-4}m$

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Therefore, the spacing between the slits is $\sf{\purple{3.44\times 10^{-4}m}}$