In the laser range-finding experiments of Example 17.10, the laser beam fired toward the moon spreads out as it travels because it diffracts through a circular exit as it leaves the laser. In order for the reflected light to be bright enough to detect, the laser spot on the moon must be no more than 1 km in diameter. Staying within this diameter is accomplished by using a special large-diameter laser. If λ = 532 nm, what is the minimum diameter of the circular opening from which the laser beam emerges? The earth-moon distance is 384,000 km.

Question

Donna Christian

Physics

4 January, 22:49

In the laser range-finding experiments of Example 17.10, the laser beam fired toward the moon spreads out as it travels because it diffracts through a circular exit as it leaves the laser. In order for the reflected light to be bright enough to detect, the laser spot on the moon must be no more than 1 km in diameter. Staying within this diameter is accomplished by using a special large-diameter laser. If λ = 532 nm, what is the minimum diameter of the circular opening from which the laser beam emerges? The earth-moon distance is 384,000 km.

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Answers (1)

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Imani Owens · Answer 1

a = 2.5 10⁻³ m = 2.5 mm

Explanation:

The diffraction equation for a slit is

. a sin θ = m λ

In the case of circular opening when using polar coordinates, the angle sought is at the first minimum m = 1, the equation is

a sin θ = 1.22 λ

Let's look for the angle of dispersion

tan θ = y / x

y = 1 km = 1000 m

tan θ = 1000 / 3.84 10⁶ = 2.6 10⁻⁴

θ = tan⁻¹ 2.6 10⁻⁴

θ = 2.6 10⁻⁴ rad

a = 1.22 λ / sin θ

a = 1.22 532.10⁻⁹ / sin 2.6 10-4

a = 2.5 10⁻³ m