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Chapter 32: Problem 30

An object is located at a distance of \(100 . \mathrm{cm}\) from a concave mirror of focal length \(20.0 \mathrm{~cm}\). Another concave mirror of focal length \(5.00 \mathrm{~cm}\) is located \(20.0 \mathrm{~cm}\) in front of the first concave mirror. The reflecting sides of the two mirrors face each other. What is the location of the final image formed by the two mirrors and the total magnification by the combination?

Short Answer

Expert verified
The final image formed by the combination of two concave mirrors is located at infinity, and the total magnification is 0.

Step by step solution

01

Find the image formed by the first mirror

To find out the image formed by the first mirror, we'll use the mirror formula: \( \frac{1}{f} = \frac{1}{d_o}+\frac{1}{d_i}\) Here, \(f = 20.0\;\text{cm}\) (focal length of the first mirror), \(d_o = 100\;\text{cm}\) (object distance from the first mirror), and \(d_i\) is the image distance we need to find. Plug in the values: \( \frac{1}{20} = \frac{1}{100}+\frac{1}{d_i} \) Solve for \(d_i\): \( d_i = \frac{100 \times 20}{100 - 20} = \frac{2000}{80} = 25\;\text{cm}\) So, the first image is formed at a distance of 25 cm in front of the first mirror.
02

Calculate magnification for the first mirror

Now, let's find the magnification of the first mirror using the formula: \( m_1 = -\frac{d_i}{d_o}\) Plug in the values we found in the previous step: \( m_1 = -\frac{25}{100} = -0.25 \) So the magnification of the first mirror is \(-0.25\).
03

Determine the new object distance for the second mirror

Since the second mirror is located \(20\;\text{cm}\) in front of the first mirror, the distance between the first image and the second mirror is: \( d_o' = 25 - 20 = 5\;\text{cm}\) Thus, the new object distance for the second mirror is 5 cm.
04

Find the final image formed by the second mirror

Now, we'll use the mirror formula again for the second mirror: \( \frac{1}{f'} = \frac{1}{d_o'}+\frac{1}{d_i'}\) Here, \(f' = 5.00\;\text{cm}\) (focal length of the second mirror), \(d_o' = 5\;\text{cm}\) (the new object distance), and \(d_i'\) is the final image distance we want to find. Plug in the values: \( \frac{1}{5} = \frac{1}{5}+\frac{1}{d_i'} \) Solve for \(d_i'\): \( d_i' = \infty\) This implies that the final image formed by the second mirror is at infinity.
05

Calculate magnification for the second mirror

Now, let's find the magnification of the second mirror using the formula: \( m_2 = -\frac{d_i'}{d_o'}\) Since \(d_i' = \infty\), the magnification of the second mirror is zero: \( m_2 = 0 \)
06

Calculate the total magnification

Finally, to find the total magnification by the combination of the two mirrors, we need to multiply the magnifications of both mirrors: \( m_{total} = m_1 \times m_2 \) \( m_{total} = -0.25 \times 0 \) \( m_{total}=0\) In conclusion, the final image formed by the combination of the two mirrors is located at infinity, and the total magnification is 0.

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