Index of Refraction of a lens+mirror

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SUMMARY

The discussion centers on calculating the index of refraction of a converging lens with a radius of curvature of 18 cm on both sides, where one side is coated with silver to act as a mirror. The effective focal length of the lens-mirror system is 5.0 cm. Using the lens maker's equation, 1/f = (n-1)(1/R1 - 1/R2), the correct index of refraction for the lens material is determined to be 1.4. Participants clarified the misunderstanding regarding the focal length of the mirror side, emphasizing that it does not directly equate to the overall system's focal length.

PREREQUISITES
  • Understanding of the lens maker's equation
  • Knowledge of focal lengths and their relationship to curvature
  • Basic principles of optics, particularly regarding lenses and mirrors
  • Familiarity with the concept of index of refraction
NEXT STEPS
  • Study the derivation and applications of the lens maker's equation
  • Explore the principles of optical systems involving mirrors and lenses
  • Investigate the effects of different materials on the index of refraction
  • Learn about practical applications of converging lenses in optical devices
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Students in optics, physics educators, and anyone involved in designing or analyzing optical systems, particularly those incorporating lenses and mirrors.

Erubus
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Homework Statement


The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated with silver so that the inner surface is reflective. When light is incident on the uncoated side it passes through the lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of a mirror with focal length 5.0 cm. What is the index of refraction of the lens material? (Answer 1.4)


Homework Equations



\frac{1}{f}=(n-1)(\frac{1}{R1}-\frac{1}{R2})

The Attempt at a Solution


I plugging in 5cm for f and 18cm for R1. For R2 I thought that because the mirror half of the lens has a focal length of 5cm, that the radius of the the silver coated side would be 10cm. I plug this into R2 however, this is incorrect probably because I'm not clear how the radius of curvature could change for either side, when it is stated at the beginning that both sides are 18cm.
 
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because the mirror half of the lens has a focal length of 5cm, that the radius of the the silver coated side would be 10cm
You appear to have misread the question - the focal length of lens + mirror is 5cm.
The focal length for the mirror side by itself is not 5cm.
 
Got it, thanks.
 

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