Question

A biconvex lens with equal radii of curvature $R$ has a focal length $f$. If the refractive index of the lens is $\frac{3}{2}$, the focal length of the lens is

• A. $R$
• B. $\frac{R}{2}$
• C. $2R$
• D. $\frac{3}{2}R$
• E. $3R$

Hint:

For a glass biconvex lens ($\mu = 1.5$) with equal radii, the focal length is always equal to the radius of curvature ($f = R$). This is a standard result useful for competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The focal length of a thin lens in air is determined by its refractive index and the radii of curvature of its two spherical surfaces.
Step 2: Key Formula or Approach:
Use the Lens Maker's Formula:
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] For an equi-biconvex lens, the first surface curves outward (\(R_1 = +R\)) and the second surface curves inward relative to the incident light (\(R_2 = -R\)).
Step 3: Detailed Explanation:
Given values:
Refractive index \(\mu = \frac{3}{2}\)
Radii: \(R_1 = R\), \(R_2 = -R\)
Substitute into the formula:
\[ \frac{1}{f} = \left(\frac{3}{2} - 1\right)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right) \] \[ \frac{1}{f} = \left(\frac{1}{2}\right)\left(\frac{1}{R} + \frac{1}{R}\right) \] \[ \frac{1}{f} = \left(\frac{1}{2}\right)\left(\frac{2}{R}\right) \] The \(2\)'s cancel out:
\[ \frac{1}{f} = \frac{1}{R} \] Inverting both sides:
\[ f = R \] Step 4: Final Answer:
The focal length of the lens is R.