Refractive Index - Lens question

[jacksonwiley]

Homework Statement

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

Homework Equations

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

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?

 

[Vibhor]

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

 

[jacksonwiley]

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

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

 

[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.

 

[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).