A lighted candle is placed 40.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a verging lens (f = 10.0 cm) that is placed 9.00 cm from the diverging lens. The final image is real, inverted, and 35.0 beyond the converging lens. Find the focal length of the diverging lens. cm ssfo f60 s
A lighted candle is placed 40.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a verging lens (f = 10.0 cm) that is placed 9.00 cm from the diverging lens. The final image is real, inverted, and 35.0 beyond the converging lens. Find the focal length of the diverging lens. cm ssfo f60 s
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![**Physics Problem: Lens System Analysis**
**Problem Statement:**
A lighted candle is placed 40.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens with a focal length \(f = 10.0 \, \text{cm}\), which is positioned 9.00 cm from the diverging lens. The final image formed is real, inverted, and located 35.0 cm beyond the converging lens. Determine the focal length of the diverging lens.
**Solution:**
Set up the lens system using the lens formula for each lens:
1. **Diverging Lens:**
Use the lens formula:
\[
\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}
\]
where \(f_1\) is the focal length of the diverging lens, \(v_1\) is the image distance from the diverging lens, and \(u_1 = -40.0 \, \text{cm}\) (since the object is in front of the lens).
2. **Converging Lens:**
The image formed by the diverging lens acts as a virtual object for the converging lens. The object distance for the converging lens can be written as:
\[
u_2 = v_1 + 9.0 \, \text{cm}
\]
where \(v_2 = 35.0 \, \text{cm}\) from the converging lens and \(f_2 = 10.0 \, \text{cm}\).
Use the lens formula:
\[
\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}
\]
3. **Solving for \(f_1\):**
Rearrange equations to find \(v_1\) and then use it to determine \(f_1\).
By solving these equations, you can find the focal length of the diverging lens.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c6fb9be-cb3b-43fd-a6cc-90b4ac965f54%2F96692d3a-d389-4c31-a285-a391546d8cf2%2Fwwp3whe_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Lens System Analysis**
**Problem Statement:**
A lighted candle is placed 40.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens with a focal length \(f = 10.0 \, \text{cm}\), which is positioned 9.00 cm from the diverging lens. The final image formed is real, inverted, and located 35.0 cm beyond the converging lens. Determine the focal length of the diverging lens.
**Solution:**
Set up the lens system using the lens formula for each lens:
1. **Diverging Lens:**
Use the lens formula:
\[
\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}
\]
where \(f_1\) is the focal length of the diverging lens, \(v_1\) is the image distance from the diverging lens, and \(u_1 = -40.0 \, \text{cm}\) (since the object is in front of the lens).
2. **Converging Lens:**
The image formed by the diverging lens acts as a virtual object for the converging lens. The object distance for the converging lens can be written as:
\[
u_2 = v_1 + 9.0 \, \text{cm}
\]
where \(v_2 = 35.0 \, \text{cm}\) from the converging lens and \(f_2 = 10.0 \, \text{cm}\).
Use the lens formula:
\[
\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}
\]
3. **Solving for \(f_1\):**
Rearrange equations to find \(v_1\) and then use it to determine \(f_1\).
By solving these equations, you can find the focal length of the diverging lens.
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