# Index of refraction and focal length question

1. Jun 14, 2010

### element41\$

1. The problem statement, all variables and given/known data

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)?

2. Relevant equations

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

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 14, 2010

### rl.bhat

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