Light of wavelength 645 nm is incident on a double

Ответ:

igtguith

01.01.2022

Slit separation, $d=1.13\times 10^{-5}\ m$

Explanation:

It is given that,

Wavelength of light, $\lambda=645\ nm=645\times 10^{-9}\ m$

The angle between the central bright fringe and the second dark fringe, $\theta=4.9^{\circ}$

In case of double slits, the condition for dark fringe is given by :

$d\ sin\theta=(n-\dfrac{1}{2})\lambda$

n = 2

$d\ sin\theta=\dfrac{3}{2}\lambda$

$d=\dfrac{3\lambda}{2\ sin\theta}$

$d=\dfrac{3\times 645\times 10^{-9}}{2\ sin(4.9)}$

$d=1.13\times 10^{-5}\ m$

So, the slit separation is $1.13\times 10^{-5}\ m$ . Hence, this is the required solution.

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alyssaxsanchez

31.03.2020

In position A

The mirror has to be observed in the ray reflected from the mirror.

Since the the angle of reflection is equal to the angle of incidence, the reflected ray must be in equal path as the incident ray.

That phenomenon is only in diagram A

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