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Chapter 2: Problem 63

An object of height \(3.0 \mathrm{cm}\) is placed at \(25 \mathrm{cm}\) in front of a diverging lens of focal length \(20 \mathrm{cm}\). Behind the diverging lens, there is a converging lens of focal length \(20 \mathrm{cm} .\) The distance between the lenses is \(5.0 \mathrm{cm}\). Find the location and size of the final image.

Short Answer

Expert verified
The final image is located \(25.33\, cm\) behind the converging lens and has a height of \(3.20\, cm\).

Step by step solution

01

Analyze the problem

Since we have two lenses, we first need to find the image size and position created by the first lens (diverging), which will act as the object for the second lens (converging). Then, we will use the resulting object's information to find the image location and size created by the second lens.
02

Use the lens formula for the diverging lens

To find the position of the image created by the diverging lens, use the lens formula: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] Where \(f\) is the focal length, \(d_o\) is the object distance and \(d_i\) is the image distance. In this case: \(f = -20\, cm\), as the lens is diverging, and \(d_o = 25\, cm\). Plugging these values into the lens formula, we get: \[\frac{1}{-20\,\text{cm}} = \frac{1}{25\,\text{cm}} + \frac{1}{d_i}\]
03

Solve for the image distance \(d_i\) for the diverging lens

Solving the equation for \(d_i\): \[\frac{1}{d_i} = \frac{1}{-20\,\text{cm}} - \frac{1}{25\,\text{cm}}\] \[\frac{1}{d_i} = - \frac{5}{-500\,\text{cm}^2}\] \[\Rightarrow d_i = -100\,cm\] The negative sign indicates that the image is virtual and formed on the same side as the object.
04

Find the magnification for the diverging lens

The magnification formula for a lens is given by: \[M = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\] Where \(h_o\) is the object height and \(h_i\) is the image height. In this case: \(h_o = 3\,cm\), and \(d_i = -100\,cm\) from Step 3. Plugging these values into the magnification formula, we get: \[M = \frac{-(-100\,\text{cm})}{25\,\text{cm}} = 4\]
05

Calculate the image height \(h_i\) for the diverging lens

To find the image height created by the diverging lens, multiply the object height by the magnification factor: \[h_i = M \cdot h_o\] Plugging in the values obtained so far, we find: \[h_i = 4 \cdot 3\,\text{cm} = 12\,\text{cm}\] Now, the image created by the diverging lens acts as the object for the converging lens.
06

Calculate the object distance for the converging lens

The distance between the lenses is given as 5.0 cm. Since the image from the diverging lens is virtual and 100 cm in front of the lens, we have: \(d_o = 100\,cm - 5\,cm = 95\,cm\) as the object distance for the converging lens.
07

Use the lens formula for the converging lens

Now, we use the lens formula for the converging lens, which has a focal length \(f = 20\, cm\): \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] \[\frac{1}{20\,\text{cm}} = \frac{1}{95\,\text{cm}} + \frac{1}{d_i}\]
08

Solve for the image distance \(d_i\) for the converging lens

Solving for \(d_i\): \[\frac{1}{d_i} = \frac{1}{20\,\text{cm}} - \frac{1}{95\,\text{cm}}\] \[\Rightarrow d_i = \frac{1}{\frac{75}{1900\,\text{cm}^2}} = 25.33\,\text{cm}\]
09

Find the magnification for the converging lens

Using the magnification formula for the converging lens: \[M = \frac{h_i'}{h_o'} = \frac{d_i}{d_o}\] Using the object distance \(d_o = 95\,cm\), and image distance as \(d_i = 25.33\,cm\) for the converging lens, we get: \[M = \frac{25.33\,\text{cm}}{95\,\text{cm}} = 0.267\]
10

Calculate the final image height

Now, using the magnification and the object height for the converging lens (which is the image height from the diverging lens, i.e., \(h_o' = 12\,\text{cm}\)): \[h_i' = M \cdot h_o'\] \[h_i' = 0.267 \cdot 12\,\text{cm} = 3.20\,\text{cm}\] So the final image is located 25.33 cm behind the converging lens and has a height of 3.20 cm.

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