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JulsSmile [24]
3 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?
Physics
1 answer:
Novay_Z [31]3 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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Answer:

The uncertainty in momentum changes by a factor of 1/2.

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