1. The problem statement, all variables and given/known data A converging meniscus lens is made of glass with index of refraction n = 1.55, and its sides have radii of curvature of 4.5 cm and 9 cm. The concave surface is placed upward and filled with carbon tetrachloride which has index of refraction n' = 1.46. Determine the focal length of the combination of glass and carbon tetrachloride. 2. Relevant equations I think these equations are relevent. [itex] 1/f = 1/f_1 +1/f_2 [/itex] (equation 1, effective focal length) [itex] 1/f = (n/n' -1)(1/R_1 - 1/R_2) [/itex] (equation 2, lensmaker's equation) Alternatively if the thin lens is surrounded by air, then [itex] 1/f = (n -1)(1/R_1 - 1/R_2) [/itex] 3. The attempt at a solution Substituting all known values into the second equation then taking the reciprocal of it to get the value of f, I get -146cm. I think the sign is correct because these are refracting interfaces, so drawing a principle ray diagram gives the focal point on the opposite side of the outgoing light, so the focal length should be negative. But I'm not confident in the magnitude of my answer, 146cm. It doesn't seem right considering that this question is worth 8 marks. I also tried to treat the meniscus lens as two concave lenses which had an one effective focal length, and the carbon liquid stuff as another concave lens which had a radius of curvature of 9cm (equal to the R.O.C of the inner part of the meniscus lens) and its own effective focal length. Then, the effective focal length of the entire combination would be [itex] 1/f = 1/f_lens + 1/f_carbon [/itex] which is [itex] 1/f = 1/(1/2.25 + 1/4.5) + 1/4.5 [/itex] [itex] ∴ 1/f = 1/1.5 + 1/4.5 => f = 1.125cm [/itex] The magnitude of 1.125cm seems far more reasonable but I think the sign is incorrect based on the side of the outgoing light. Here's a sketch of how I imagine this to be: (am I imagining it correctly? Is one of my answers correct?) EDIT: I sketched the R1 incorrectly! I drew the line a little too far down. The line for R1 should stop at the first part of the lens.