Question

An equiconvex lens is made of glass of refractive index 1.55. If the focal length of the lens is 15.0 cm, calculate the radius of curvature of its surfaces.

Hint:

In an equiconvex lens, the radius of curvature on both sides is equal in magnitude but opposite in sign: \( R_1 = R, R_2 = -R \). Use the lens maker’s formula carefully with signs.

Answer 1

Given: Refractive index \( \mu = 1.55 \) Focal length \( f = 15 \ \text{cm} \) For an equiconvex lens: Radius of curvature \( R_1 = R \), \( R_2 = -R \) Apply the lens maker’s formula: \[ \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Substitute the given values: \[ \frac{1}{15} = (1.55 - 1)\left( \frac{1}{R} - \frac{1}{-R} \right) \] \[ \Rightarrow \frac{1}{15} = 0.55 \left( \frac{2}{R} \right) \] \[ \Rightarrow \frac{1}{15} = \frac{1.1}{R} \] Solve for \( R \): \[ R = 1.1 \times 15 = 16.5 \ \text{cm} \] Final Answer: Radius of curvature \( = 16.5 \ \text{cm} \)