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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

The objective lens in a laboratory microscope has a focal length of \(3.00 \mathrm{~cm}\) and provides an overall magnification of \(1.0 \cdot 10^{2} .\) What is the focal length of the eyepiece if the distance between the two lenses is \(30.0 \mathrm{~cm}\) ?

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?

Several small drops of paint (less than \(1 \mathrm{~mm}\) in diameter) splatter on a painter's eyeglasses, which are approximately \(2 \mathrm{~cm}\) in front of the painter's eyes. Do the dots appear in what the painter sees? How do the dots affect what the painter sees?

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?

An object is \(6.0 \mathrm{~cm}\) from a converging thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm}\), determine the image magnification.
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