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Chapter 33: Problem 31

As a high-power laser engineer you need to focus a 1.06-mm diameter laser beam to a 10.0 - \(\mu\) m diameter spot \(20.0 \mathrm{~cm}\) behind the lens. What focal length lens would you use?

Short Answer

Expert verified
Answer: The required focal length for the lens is approximately 19.95 mm or 19952.12 μm.

Step by step solution

01

Write down the given information and convert into same units.

We have the following given information: - Diameter of the laser beam (object size): 1.06 mm or 1060 \(\mu\)m. - Diameter of the focused spot (image size): 10.0 \(\mu\)m. - Distance behind the lens (image distance): 20.0 cm or 20000 \(\mu\)m.
02

Calculate the magnification.

Magnification is the ratio of the size of the image to the size of the object. The formula for magnification is: $$ M = \frac{h_i}{h_o} $$ Where \(M\) is the magnification, \(h_i\) is the image size or focused spot diameter, and h_o is the object size or laser beam diameter. Plug in the values we know: $$ M =\frac{10.0 \mu m}{1060 \mu m} = \frac{1}{106} $$
03

Calculate object distance.

The magnification formula also relates the distances from the lens to the image and object: $$ M = -\frac{d_i}{d_o} $$ Where \(M\) is the magnification, \(d_i\) is the image distance or the 20000 \(\mu\)m, and \(d_o\) is the object distance we're trying to find. Since we know the magnification from step 2, we have : $$ \frac{1}{106} = -\frac{20000 \mu m}{d_o} $$ Now, solve for the object distance, \(d_o\): $$ d_o = -20000 \mu m \cdot 106 = -2120000 \mu m $$ The negative sign indicates that the object is on the same side of the lens as the image, which is typical for laser focusing systems.
04

Use the lens formula to find the focal length.

The lens formula is given by: $$ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $$ Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. We can plug in the known values and solve for the focal length, \(f\): $$ \frac{1}{f} = \frac{1}{-2120000 \mu m} + \frac{1}{20000 \mu m} $$ Now, solve for \(f\): $$ f = \frac{1}{\frac{1}{-2120000 \mu m} + \frac{1}{20000 \mu m}} \approx 19952.12 \mu m $$ So, the required focal length lens is approximately 19.95 mm or 19952.12 μm.

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Most popular questions from this chapter

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?

Some reflecting telescope mirrors utilize a rotating tub of mercury to produce a large parabolic surface. If the tub is rotating on its axis with an angular frequency \(\omega,\) show that the focal length of the resulting mirror is: \(f=g / 2 \omega^{2}\).

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