Question

A biconvex lens ($R_1 = R_2 = 30\ \text{cm}$) has focal length equal to the focal length of a concave mirror. The radius of curvature of the concave mirror is [Refractive index of material of lens $= 1.6$]

• A. $30\ \text{cm}$
• B. $40\ \text{cm}$
• C. $50\ \text{cm}$
• D. $20\ \text{cm}$

Hint:

For any symmetric biconvex lens where $R_1 = R_2 = R$, the Lens Maker's Formula simplifies directly to $f = \frac{R}{2(\mu - 1)}$. Substituting $R = 30$ and $\mu = 1.6$ gives $f = \frac{30}{2(0.6)} = \frac{30}{1.2} = 25\ \text{cm}$ in one quick step!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the focal length of a symmetric biconvex lens where both radii of curvature are identical.

Step 2: Key Formula or Approach:
The Lens Maker's Formula: 1/f = (μ - 1)(1/R₁ - 1/R₂). For a symmetric lens with R₁ = R and R₂ = -R, this simplifies to 1/f = (μ - 1)(2/R), giving f = R/[2(μ - 1)].

Step 3: Detailed Explanation:
Substituting R = 30 cm and μ = 1.6 into the simplified expression: f = 30/[2(1.6 - 1)] = 30/[2(0.6)] = 30/1.2 = 25 cm. The symmetry condition collapses the formula into a single quick computation without sign convention complications.

Step 4: Final Answer:
The focal length is 25 cm.