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2 years ago

5

A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through

the telescope in opposite the normal direction and can then be projected onto a satellite or the Moon. If this is done with the Hale telescope, producing a 5.08 m diameter beam of 613 nm light, what is the minimum angular spread of the beam?

1 answer:

Novay_Z [31]2 years ago

5 0

Answer:

Angular spread, $\theta=1.472\times 10^{-7}\ rad$

Explanation:

It is given that,

Wavelength of the light, $\lambda=613\ nm=613\times 10^{-9}\ m$

Diameter of the telescope, D = 5.08 m

The minimum angular spread is given by :

$\theta=\dfrac{1.22\lambda}{D}$

$\theta=\dfrac{1.22\times 613\times 10^{-9}}{5.08}$

$\theta=1.472\times 10^{-7}\ rad$

So, the minimum angular spread of the beam is $1.472\times 10^{-7}\ radian$ . Hence, this is the required solution.

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