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A biconvex lens has radii of curvature R1 = 10 cm (first surface) and R2 = 20 cm (second surface), and its refractive index is mu = 1.5. The refractive indices of the medium on the left and right sides of the lens are mu1 and mu2 respectively. A bundle of parallel rays enters from the left. Find the image distance f2, measured from the optical centre of the lens, where the rays converge.
- 20*mu1 / [4.5 - (2*mu1 + mu2)]
- 20*mu2 / [4.5 - (mu1 + 2*mu2)]
- 20*mu2 / [4.5 - (2*mu1 + mu2)]
- 20*mu1 / [4.5 - (mu1 + 2*mu2)]
Correct answer: 20*mu2 / [4.5 - (mu1 + 2*mu2)]
Solution
Using refraction at each surface with the standard formula, and taking R1 = +10 cm and R2 = -20 cm (sign convention for biconvex lens), applying refraction at surface 1 with object at infinity, then using the result as input for surface 2, yields f2 = 20*mu2 / [4.5 - (mu1 + 2*mu2)].
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