Compute the focal length in air of a thin biconvex lens

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SUMMARY

The focal length of a thin biconvex lens with a refractive index of 1.5 and radii of 20 cm and 40 cm is calculated using the lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2), resulting in a focal length of 80 cm. When an object is placed 40 cm from the lens, the magnification (Mt) is calculated as -f/x0, yielding a value of -2 cm, indicating an inverted image. The calculations confirm that the image is virtual and magnified by a factor of 2.

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  • Understanding of the lens maker's formula
  • Knowledge of optical properties of biconvex lenses
  • Familiarity with concepts of magnification and image formation
  • Basic principles of geometric optics
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jlmac2001
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Compute the focal length in air of a thin biconvex lens (n=1.5) having radii of 20 and 40 cm. Locate and describe the image of an object 40cm from the lens.

my answer:

1/f = (n-1)(1/R1-1/R2)
1/f = (1.5-1)(1/20-1/40)
1/f = .5(.025)
f = 1/.0125 = 80cm (convering lens because f is positive)

Mt=-f/x0=-80/40=-2cm (Inverted image because Mt is negative)

Did I use the correct equations?
 
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Originally posted by jlmac2001
f = 1/.0125 = 80cm
Looks O.K. to me.
Mt=-f/x0=-80/40=-2cm
I should expect, if the object is only 40cm from the lens while the lense's focal length is 80cm, then the image should be upright and virtual, and magnified by a factor of 2.
 


Yes, you used the correct equations to calculate the focal length and image distance for a thin biconvex lens. Your calculations are also correct, and your description of the image is accurate. Well done!
 

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