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.
1
Expert's answer
2018-07-30T14:06:09-0400
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
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