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Chapter 33: Problem 44

Two converging lenses with focal lengths \(5.00 \mathrm{~cm}\) and \(10.0 \mathrm{~cm}\), respectively, are placed \(30.0 \mathrm{~cm}\) apart. An object of height \(h=5.00 \mathrm{~cm}\) is placed \(10.0 \mathrm{~cm}\) to the left of the \(5.00-\mathrm{cm}\) lens. What will be the position and height of the final image produced by this lens system?

Short Answer

Expert verified
Answer: The final image is located at a distance of 50.0 cm to the right of the second lens and has a height of 5.00 cm.

Step by step solution

01

Define symbols and given values

We will use the following symbols for the given values: \(f_1 = 5.00cm\) (the focal length of the first lens) \(f_2 = 10.0cm\) (the focal length of the second lens) \(d_{12} = 30.0cm\) (the distance between the two lenses) \(d_o = 10.0cm\) (the object's distance from the first lens) \(h_o = 5.00cm\) (the height of the object) We need to find the position \(d_i\) and height \(h_i\) of the final image.
02

Apply the lens formula for the first lens

We can calculate the image distance produced by the first lens using the lens formula: \(\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_{i_1}}\) Now, we can solve for \(d_{i_1}\): \(d_{i_1} = \frac{1}{\frac{1}{f_1} - \frac{1}{d_o}}\) Substitute the values and calculate \(d_{i_1}\): \(d_{i_1} = \frac{1}{\frac{1}{5} - \frac{1}{10}} = 10cm\)
03

Calculate the magnification for the first lens

Next, we find the magnification produced by the first lens using the magnification formula: \(m_1 = -\frac{d_{i_1}}{d_o}\) Substitute the values and calculate \(m_1\): \(m_1 = -\frac{10}{10} = -1\)
04

Determine the intermediate object distance for the second lens

Now, we need to find the distance of the intermediate object (formed by the first lens) from the second lens. \(d_{o_2} = d_{12} - d_{i_1} = 30 - 10 = 20cm\)
05

Apply the lens formula for the second lens

We can calculate the image distance produced by the second lens using the lens formula: \(\frac{1}{f_2} = \frac{1}{d_{o_2}} + \frac{1}{d_{i_2}}\) Now, we can solve for \(d_{i_2}\): \(d_{i_2} = \frac{1}{\frac{1}{f_2} - \frac{1}{d_{o_2}}}\) Substitute the values and calculate \(d_{i_2}\): \(d_{i_2} = \frac{1}{\frac{1}{10} - \frac{1}{20}} = 20cm\)
06

Calculate the total image distance from the second lens

The total image distance from the second lens is just the sum of the distances from the first lens and the image produced by the second lens: \(d_i = d_{12} + d_{i_2} = 30.0 + 20.0 = 50.0cm\)
07

Calculate the magnification for the second lens

Next, we find the magnification produced by the second lens using the magnification formula: \(m_2 = -\frac{d_{i_2}}{d_{o_2}}\) Substitute the values and calculate \(m_2\): \(m_2 = -\frac{20}{20} = -1\)
08

Calculate the overall magnification

The overall magnification of the system is the product of magnifications of the individual lenses: \(m = m_1 \times m_2 = (-1) \times (-1) = 1\)
09

Calculate the final image height

Lastly, we can find the height of the final image using the overall magnification: \(h_i = m \times h_o = 1 \times 5.00 = 5.00cm\)
10

Interpret the results

The final image produced by the lens system is located at a distance of \(50.0 cm\) to the right of the second lens and has a height of \(5.00 cm\). Since the overall magnification is positive, the image is upright and the same size as the object.

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Most popular questions from this chapter

A refracting telescope has the objective lens of focal length \(10.0 \mathrm{~m}\). Assume it is used with an eyepiece of focal length \(2.00 \mathrm{~cm}\). What is the magnification of this telescope?

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?

An instructor wants to use a lens to project a real image of a light bulb onto a screen \(1.71 \mathrm{~m}\) from the bulb. In order to get the image to be twice as large as the bulb, what focal length lens will be needed?

An object is moved from a distance of \(30 \mathrm{~cm}\) to a distance of \(10 \mathrm{~cm}\) in front of a converging lens of focal length \(20 \mathrm{~cm}\). What happens to the image? a) Image goes from real and upright to real and inverted. b) Image goes from virtual and upright to real and inverted. c) Image goes from virtual and inverted to real and upright. d) Image goes from real and inverted to virtual and upright. e) None of the above.

An unknown lens forms an image of an object that is \(24 \mathrm{~cm}\) away from the lens, inverted, and a factor of 4 larger in size than the object. Where is the object located? a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens c) \(96 \mathrm{~cm}\) from the lens on the same side of the lens d) \(96 \mathrm{~cm}\) from the lens on the other side of the lens e) No object could have formed this image.
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