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Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?

Short Answer

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An object is placed \(2.50 \mathrm{m}\) to the left of the lens combination. Answer: (a) The image will be located at \(5.10 \mathrm{cm}\) to the left of the lens combination. (b) The image is virtual.

Step by step solution

01

Find the equivalent focal length of the lens combination

To find the equivalent focal length of the lens combination, we'll use the formula for the combined focal length of two thin lenses with focal lengths \(f_1\) and \(f_2\) which are placed very close together: \(1/f_{eq} = 1/f_{1} + 1/f_{2}\) Given that \(f_{1} = 4.00 \mathrm{cm}\) (converging lens) and \(f_{2} = -20.0\mathrm{cm}\) (diverging lens), let's plug in these values in the formula. \(1/f_{eq} = \frac{1}{4.00} - \frac{1}{20.0}\) Calculate \(1/f_{eq}\): \(1/f_{eq} = 0.25 - 0.05 = 0.20 \mathrm{cm^{-1}}\) Now, find out \(f_{eq}\): \(f_{eq} = 1 / 0.20 = 5.00 \mathrm{cm}\) Hence, the equivalent focal length of the lens combination is \(5.00 \mathrm{cm}\).
02

Calculate the image location and determine if it's real or virtual

We will now use the lens formula to find out the position of the image formed by the lens combination. The lens formula is given by: \(1/f_{eq} = 1/u + 1/v\) Where \(u\) is the object distance, \(v\) is the image distance, and \(f_{eq}\) is the equivalent focal length that we found in step 1 (\(5.00 \mathrm{cm}\)). Note that we're given \(u = 2.50 \mathrm{m} = 250 \mathrm{cm}\) (as the object is to the left of the lenses). Now, substitute the given values in the lens formula and solve for \(v\): \(1/5.00 = 1/250 - 1/v\) Solve for \(1/v\): \(1/v = 1/5.00 - 1/250 = 0.20 - 0.004 = 0.196 \mathrm{cm^{-1}}\) Calculate \(v\): \(v = 1 / 0.196 \approx 5.10 \mathrm{cm}\) The image is formed at \(5.10 \mathrm{cm}\) from the lens combination on the same side as the object (since the calculated \(v\) is positive). Now, let's determine if the image is real or virtual. As the lenses are converging (as the equivalent focal length is positive), the image will be real if it's formed on the other side of the lenses and virtual if it's formed on the same side. Since the image is formed on the same side as the object (to the left of the lenses), the image is virtual. In conclusion: (a) The image will be located at \(5.10 \mathrm{cm}\) to the left of the lens combination. (b) The image is virtual.

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Most popular questions from this chapter

An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.
Show that if two thin lenses are close together \((s,\) the distance between the lenses, is negligibly small), the two lenses can be replaced by a single equivalent lens with focal length \(f_{\mathrm{eq}} .\) Find the value of \(f_{\mathrm{eq}}\) in terms of \(f_{1}\) and \(f_{2}.\)
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of \(25 \mathrm{cm} .\) The focal length of the magnifying glass is \(+5.0 \mathrm{cm} .\) Assume that Keesha's eye is close to the magnifying glass. (a) What should be the distance between the magnifying glass and the beetle? (b) What is the angular magnification? (tutorial: magnifying glass II).
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
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