Destroying a Building Using a Lens

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SUMMARY

The discussion centers around the calculations needed to determine the radius of curvature for a plano-convex lens intended to focus sunlight onto a target 60 meters away. The lens has a focal length of 60 m and an index of refraction of 1.5. The initial calculation incorrectly yielded a radius of curvature (R2) of 30 m. The relevant equations for this optical problem include the lens maker's equation and the relationship between focal length and curvature.

PREREQUISITES
  • Understanding of plano-convex lens optics
  • Familiarity with the lens maker's equation
  • Knowledge of focal length and radius of curvature relationships
  • Basic principles of light refraction and index of refraction
NEXT STEPS
  • Study the lens maker's equation in detail
  • Explore the effects of different indices of refraction on lens performance
  • Research real-world incidents involving sunlight focusing through glass
  • Investigate the design and safety considerations for architectural glass facades
USEFUL FOR

Physics students, optical engineers, architects, and safety professionals interested in the implications of lens design and sunlight focusing.

ab200
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Homework Statement
I want to mount a giant burning lens on the new engineering building and melt down the student commons. The lens will sit 60 m away from the commons (laterally) and tower 40 m above its roof. I can only afford to grind one side of the giant lens to be spherical, so the other side will have to be flat. The glass will have an index of 1.5. What radius of curvature in m is needed to focus the suns rays onto the commons?
Relevant Equations
(1/s) + (1/s') = 1/f = (n-1)[(1/R1) - (1/R2)]
(n1/s) + (n2/s') = (n2 - n1)/R
Putting the questionable nature of my professor's problem-writing aside, I found this question pretty confusing. I assumed that 60 m was the focal length, and that R1 was infinity since it is a flat surface. This gives me:
1/60 = (1.5 - 1)[(1/∞) - (1/R2)]
This gave me R2 = 30 m, but that isn't correct.
 
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ab200 said:
Homework Statement: I want to mount a giant burning lens on the new engineering building and melt down the student commons. The lens will sit 60 m away from the commons (laterally) and tower 40 m above its roof. I can only afford to grind one side of the giant lens to be spherical, so the other side will have to be flat. The glass will have an index of 1.5. What radius of curvature in m is needed to focus the suns rays onto the commons?
Relevant Equations: (1/s) + (1/s') = 1/f = (n-1)[(1/R1) - (1/R2)]
(n1/s) + (n2/s') = (n2 - n1)/R

Putting the questionable nature of my professor's problem-writing aside, I found this question pretty confusing. I assumed that 60 m was the focal length, and that R1 was infinity since it is a flat surface. This gives me:
1/60 = (1.5 - 1)[(1/∞) - (1/R2)]
This gave me R2 = 30 m, but that isn't correct.
What is distance?
 
@hutchphd's queston is well worth considering but there is an additional issue with your answer.
ab200 said:
1/60 = (1.5 - 1)[(1/∞) - (1/R2)]
In terms of symbols this becomes

1/s=-(n-1)/R

Is this the correct equation for a plano-convex lens? Look it up.
 
hutchphd said:
What is distance?
I got it! Thanks for the tip.
 
FWIW, I've read of fires started by sunlight focussed onto adjacent, readily flammable material by the curve of a vase or clear bottle, even the dished base of a drinks can. IIRC, the latter, 'merely' thoughtlessly tossed roadside trash, was determined to be the surprising source of a massive 'bush-fire'...

Let's not forget architectural howlers such as the US and UK skyscrapers whose curved glazed facades did their best to 'Death_Star Zap' their neighbourhood until re-fitted with louvred shades...
 
Nik_2213 said:
I've read of fires started by sunlight focussed
Please link to a credible source. You've been here long enough to know that "I've read somewhere" is not a valid source for PF. Thanks.
 
This poor MacGregor owner almost had his boat burn to the pavement from a warped mirror in the head*!



* bathroom, for you landlubbers
 

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