Answer to Question #79415 in Optics for Khadijah

Question

The magnitude of the lens’ radii of curva- ture are R1 = 2.616 cm and R2 = 1.962 cm.
Assume: The lens is made of glass whose index of refraction is 1.51.
Note: This is the same lens that is shown in the figure above.
What is the focal length of this lens? Answer in units of cm.

Answer 1

The focal length f of a thin lens made of material with refractive index n, and surfaces with radii of curvature R_1 and R_2 is given by
1/f=(n-1)(1/R_1 +1/R_2 )
Since n=1.51, R_1=1.616 cm and R_2=1.962 cm, we obtain
1/f=(1.51-1)(1/(1.616 cm)+1/(1.962 cm))
f=1.738 cm