Radius of curvature of second lens surface

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SUMMARY

The discussion centers on calculating the radius of curvature for the second surface of a lens with a power of -5.0D and a convex first surface radius of 15.0cm, made from a material with a refractive index of 1.50. The initial calculation yielded an incorrect radius of -0.10cm, while the correct answer is -6.0cm. The error arose from not converting units properly, as the focal length must be in meters, while the radius was mistakenly used in centimeters.

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  • Understanding of lens power and its relationship to focal length
  • Knowledge of the lens maker's formula
  • Familiarity with unit conversions, specifically between centimeters and meters
  • Basic principles of optics, including refractive index
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  • Explore the implications of refractive index on lens design
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Students studying optics, physics educators, and anyone involved in lens design or optical engineering will benefit from this discussion.

desmond iking
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Homework Statement



A lens of power of -5.0D has a surface which is convex of radius of curvature of 15.0cm. The lens is made of material of refractive index of 1.50.
What's the radius of the other surface of lens?


Homework Equations





The Attempt at a Solution


since power = 1/f
so i have (1/ (-1/5) )= ( (1.5-1)/ 1) x ( (1/15) + (1/r2) )
i found out my r2 = -0.10cm.
But the correct ans given is -6.0cm.
 
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Here, lens of power = -5.0D. (1 Diopter = 1m-1)

And you have taken the P = 1/f, here f is in "meter", but you have taken R1 = 15 cm which is in "cm."

Don't you think this is weird.
 

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