Geometric Optics: A Diverging and Converging Lense in Series

In summary: Therefore, the focal length of the diverging lens is 14.09 cm. In summary, the romantic, lighted candle is placed 39 cm in front of a diverging lens and passes through a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 21 cm beyond the converging lens. The focal length of the diverging lens is 14.09 cm.
  • #1
cwatki14
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A romantic, lighted candle is placed 39 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 21 cm beyond the converging lens. Find the focal length of the diverging lens.

-1/S + 1/S' = 1/f
Here's my approach which led me to the wrong answer:
I started by just looking a the converging lense. The image's location is 21 to the right of the lense, or at +21cm. The primary focal point of the converging lense should be to the left, or at -10cm. Here we solve for S using the equation I listed about.
-1/S=1/f - 1/S'
-1/S= 1/10 - 1/21
So S= -19.09cm

Then I take into account that the diverging and converging lense are 5 cm apart. So I am going to subtract 5 from the S solved about.
So S' for the diverging lense= 14.090cm. Since I have to create a new coordinate system, this image is to the left of the diverging lense, so it is at -14.090cm.
Then I use -1/S + 1/S' = 1/f again to solve for f.
so -1/-39cm + 1/-14.090cm= 1/f
I solved for f, and got f= -22.0598, which makes sense because the image should be closer than the focal point, but it's wrong...

Any ideas?
 
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  • #2
The equation you used is incorrect. The correct equation is 1/f = 1/S + 1/S'. Using this equation, we have:1/f = 1/(39 cm) + 1/(-14.09 cm)1/f = -1/14.09 cmf = 14.09 cm
 
  • #3


I would first check my calculations to make sure there were no errors. I would also double check the given information to ensure that all the values are correct. If everything checks out, I would then consider the possibility of the lenses being mislabeled or misaligned. It is possible that the diverging lens is actually a converging lens and vice versa. I would also consider the possibility of any other factors, such as refraction or diffraction, that may affect the final image. If all else fails, I would conduct experiments to replicate the scenario and see if I can arrive at the given solution.
 

1. What is the difference between a diverging and converging lens?

A diverging lens is a lens that causes parallel rays of light to spread out or diverge, while a converging lens is a lens that causes parallel rays of light to come together or converge.

2. How do a diverging and converging lens work together in series?

When a diverging and converging lens are placed in series, the diverging lens first causes the light rays to spread out, and then the converging lens brings the rays back together, resulting in a focused image.

3. Can a diverging and converging lens be used separately?

Yes, a diverging and converging lens can be used separately. A diverging lens can be used to correct nearsightedness, while a converging lens can be used to correct farsightedness.

4. What are some real-life applications of geometric optics with diverging and converging lenses?

Geometric optics with diverging and converging lenses are used in eyeglasses, cameras, telescopes, and microscopes to correct vision and magnify images.

5. How are the properties of a diverging and converging lens affected by the curvature of the lens?

The curvature of the lens affects the focal length and power of the lens. A lens with a more curved surface will have a shorter focal length and a higher power, while a lens with a less curved surface will have a longer focal length and a lower power.

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