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High-power lasers are used to cut and weld materials by focusing the laser beam to a very small spot. This is like using a magnifying lens to focus the sun’s light to a small spot that can burn things. As an engineer, you have designed a laser cutting device in which the material to be cut is placed 5.0 cm behind the lens. You have selected a high-power laser with a wavelength of 1.06 mm. Your calculations indicate that the laser must be focused to a 5.0@mm@diameter spot in order to have sufficient power to make the cut. What is the minimum diameter of the lens you must install?

Short Answer

Expert verified

This is like using a magnifying lens to focus the sun’s light to a small spot that can burn things

Step by step solution

01

Given Information

This is like using a magnifying lens to focus the sun’s light to a small spot that can burn things

02

 Step 2 : Calculation

The laser must be focused to a 5.0@mm@diameter spot in order to have sufficient power to make the cut

03

simplify

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a. For parallel light rays coming from the left, show that the effective focal length of this two-lens system is f=R/12n2-n1-12, where localid="1648757054673" n1and n2are, respectively, the indices of refraction of the diverging and the converging lenses. Don’t forget to make the thin-lens approximation.

b. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define n1 and n2 as nblue-nred for the two lenses. What value of the ratio n1/n2makes fblue=fredfor the two-lens system? That is, the two-lens system does not exhibit chromatic aberration.

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nblue nred

Crown glass 1.525 1.517

Flint glass 1.632 1.616

d. What value of R gives a focal length of 10.0cm?

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