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 lens is determined by its refractive index and the radii of curvature of its surfaces, as defined by the Lens Maker's Formula.
Key Formula or Approach:
Lens Maker's Formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Step 2: Detailed Explanation:
For a biconvex lens with equal radii:
1. $R_1 = +R$ (first surface is convex towards incident light).
2. $R_2 = -R$ (second surface is concave towards incident light).
3. Refractive index $\mu = \frac{3}{2}$.
Substituting these values into the formula:
\[ \frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
\[ \frac{1}{f} = \left( \frac{1}{2} \right) \left( \frac{1}{R} + \frac{1}{R} \right) \]
\[ \frac{1}{f} = \frac{1}{2} \times \frac{2}{R} \]
\[ \frac{1}{f} = \frac{1}{R} \implies f = R \]

Step 3: Final Answer:
The focal length of the lens is $R$.