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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

Three converging lenses of focal length \(5.0 \mathrm{~cm}\) are arranged with a spacing of \(2.0 \cdot 10^{1} \mathrm{~cm}\) between them, and are used to image an insect \(2.0 \cdot 10^{1} \mathrm{~cm}\) away. a) Where is the image? b) Is it real or virtual? c) Is it upright or inverted?

Two distant stars are separated by an angle of 35 arcseconds. If you have a refracting telescope whose objective lens focal length is \(3.5 \mathrm{~m}\), what focal length eyepiece do you need in order to observe the stars as though they were separated by 35 arcminutes?

When performing optical spectroscopy (for example, photoluminescence or Raman spectroscopy), a laser beam is focused on the sample to be investigated by means of a lens having a focal distance \(f\). Assume that the laser beam exits a pupil \(D_{o}\) in diameter that is located at a distance \(d_{\mathrm{o}}\) from the focusing lens. For the case when the image of the exit pupil forms on the sample, calculate a) at what distance \(d_{\mathrm{i}}\) from the lens is the sample located and b) the diameter \(D_{i}\) of the laser spot (image of the exit pupil) on the sample. c) What are the numerical results for: \(f=10.0 \mathrm{~cm},\) \(D_{o}=2.00 \mathrm{~mm}, d_{\mathrm{o}}=1.50 \mathrm{~m} ?\)

A person with a near-point distance of \(24.0 \mathrm{~cm}\) finds that a magnifying glass gives an angular magnification that is 1.25 times larger when the image of the magnifier is at the near point than when the image is at infinity. What is the focal length of the magnifying glass?

What is the focal length of a magnifying glass if a \(1.0-\mathrm{mm}\) object appears to be \(10 . \mathrm{mm} ?\)
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