Final answer:
The focal length of the contact lens, with an index of refraction of 1.45 and radii of curvature of +2.08 cm and +2.42 cm, is approximately 32.92 cm, indicating it is a converging lens.
Step-by-step explanation:
To find the focal length of a contact lens made of plastic with an index of refraction of 1.45, which has an outer radius of curvature of +2.08 cm and an inner radius of curvature of +2.42 cm, we can use the Lensmaker's Equation:
Lensmaker's Equation: \( \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Where:
- n is the index of refraction of the material (1.45 in this case).
- R1 is the radius of curvature of the first lens surface (+2.08 cm).
- R2 is the radius of curvature of the second lens surface (+2.42 cm).
- f is the focal length we want to find.
Plugging in the values, we get:
\( \frac{1}{f} = (1.45 - 1)\left(\frac{1}{+2.08} - \frac{1}{+2.42}\right) \)
\( \frac{1}{f} = 0.45\left(\frac{1}{+2.08} - \frac{1}{+2.42}\right) \)
\( \frac{1}{f} = 0.45\left(\frac{1}{+2.08} - \frac{1}{+2.42}\right) \)
\( \frac{1}{f} = 0.45\left(\frac{2.42 - 2.08}{(2.08)(2.42)}\right) \)
\( \frac{1}{f} = 0.45\left(\frac{0.34}{5.0336}\right) \)
\( \frac{1}{f} = 0.45\left(0.067529 \right) \)
\( \frac{1}{f} = 0.0303881 \)
Thus, the focal length f is given by:
\( f = \frac{1}{0.0303881} \)
The calculated value results in a focal length of approximately
\( f \approx 32.92 \text{ cm} \)
Note: Since the calculated focal length is positive, this indicates that the contact lens is a converging lens.