Final answer:
Using the Lensmaker's Formula, the focal length of the meniscus lens is calculated as 120 cm, which is not an option provided in the question.
Step-by-step explanation:
To find the focal length of a meniscus lens, we can use the Lensmaker's Formula:
\(\frac{1}{f} = (n - 1)(\frac{1}{R_1} - \frac{1}{R_2})\)
Where \(f\) is the focal length of the lens, \(n\) is the index of refraction of the lens material, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces. Given that \(R_1 = 20 cm\) and \(R_2 = 15 cm\), and the index of refraction \(n = 1.5\):
\(\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20 cm} - \frac{1}{15 cm}\right)\)
\(\frac{1}{f} = 0.5 \left(\frac{15 - 20}{20 \times 15}\right)\)
\(\frac{1}{f} = 0.5 \left(\frac{-5}{300}\right)\)
\(\frac{1}{f} = -\frac{5}{600}\)
\(\frac{1}{f} = -\frac{1}{120}\), or \(\frac{1}{f} = -0.0083\ cm^{-1}\)
Taking the reciprocal to find focal length, \(f = -120 cm\).
The negative sign indicates that it's a diverging lens. However, since the sign is not required for this particular question, we are only interested in the magnitude of the focal length. Thus, the focal length of the meniscus lens is 120 cm, which is not among the provided options.