# Refractive Index - Lens question!

1. Jul 1, 2014

### jacksonwiley

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

A double concave lens has equal radii of curvature of 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.0 cm from the lens. What is the index of refraction of the lens material?

A. 1.98
B. 1.9
C. 1.84
D. 2.06

2. Relevant equations

1/f = (n – 1 ) (1/R1 - 1/R2 )
1/f = 1/do + 1/di

3. The attempt at a solution

1/f = 1/14.2 + 1/5.0 so i get 1/f = 0.27

1/f = (n – 1 ) (1/R1 - 1/R2 )
0.27 = (n-1) (1/15.1 - (-1/15.1)

this results in an n equal to about 3.03 which is not close to any of the answers, though I'm sure i have the right equations, I'm just not sure what i'm doing wrong?
are these only applicable to convex lens?

Last edited: Jul 1, 2014
2. Jul 1, 2014

### Vibhor

I am getting n=1.98 .Is this the correct answer ?

3. Jul 1, 2014

### jacksonwiley

i'm not sure if that is right but it is an answer choice. what method did you use?

4. Jul 1, 2014

### Vibhor

The problem is with the calculation of focal length . You should be very careful with whatever sign convention you are using . 1/f is coming out to be -0.13 cm . I am using new coordinate sign convention where the direction of incident rays is considered positive .In this sign convention both the object as well as image distance is negative.

5. Jul 2, 2014

### dauto

If the image is virtual than di is negative (according with the most common sign convention which seems to be the one used here).

Last edited: Jul 2, 2014