Index of refraction and focal length question

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SUMMARY

The discussion focuses on calculating the index of refraction required for a lens with a focal length of 30 cm in air, using the lens maker's equation. The correct formula is identified as 1/f = (n - 1)(1/R1 + 1/R2), where R1 and R2 are the radii of curvature. The solution reveals that the index of refraction (n) must be 1.6 to achieve the desired focal length. Additionally, when the lens is placed in water (n=4/3), the focal length is calculated to be 15 cm.

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  • Understanding of the lens maker's equation
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  • Familiarity with the concept of radii of curvature
  • Basic principles of optics and refraction
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element41$
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Homework Statement



A Lens maker desires to make a lens of focal length 30 cm. in air. The equipment in the shop is set to make lens faces of 20 cm.

a. What index of refraction should the lens material have to produce this focal length?


b. What would the focal length of this lens be in water (n=4/3)?


Homework Equations



n{out}/f=(n{lens}-n{out})(1/R{1}-1/R{2}


The Attempt at a Solution



This is what i did:
a.
1/f=(n)(1/R)
1/300=(n)(1/20)
0.6=n

b.
1/f=n(1/R)
=(4/3)(0.05)
066
f=1/0.06
f=15cm

Thank you
 
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element41$ said:

Homework Statement



A Lens maker desires to make a lens of focal length 30 cm. in air. The equipment in the shop is set to make lens faces of 20 cm.

a. What index of refraction should the lens material have to produce this focal length?b. What would the focal length of this lens be in water (n=4/3)?

Homework Equations



n{out}/f=(n{lens}-n{out})(1/R{1}-1/R{2}

The Attempt at a Solution



This is what i did:
a.
1/f=(n)(1/R)

You have not used the relevant equation properly. In air it should be

\frac{1}{f} = (n-1)(\frac{1}{R_1} + \frac{1}{R_2})

For biconvex R1 = R2
 

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