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Karolina [17]
3 years ago
10

In a double-slit experiment, the second-order bright fringe is observed at an angle of 0.61°. If the slit separation is 0.11 mm,

then what is the wavelength of the light?
Physics
1 answer:
tankabanditka [31]3 years ago
8 0

Answer:

5.86\times 10^{-7}\ \text{m}

Explanation:

d = Slit separation = 0.11 mm

\theta = Angle = 0.61^{\circ}

m = Order = 2

\lambda = Wavelength

We have the relation

d\sin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{d\sin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.11\times 10^{-3}\times \sin0.61^{\circ}}{2}\\\Rightarrow \lambda=5.86\times 10^{-7}\ \text{m}

The wavelength of the light is 5.86\times 10^{-7}\ \text{m}.

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When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

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What is the difference between mutual flux, leakage flux and magnetizing flux<br> ​
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