Question

A double convex lens has radius of curvature R of each surface and refractive index of its material is \(\mu = 1.5\). We have

• A. f = R/2
• B. f = R
• C. f = -R
• D. f = 2R

Hint:

Always be careful with the sign convention in optics problems. For radii of curvature, a common convention is to measure distances from the optical center. If the center of curvature is on the side where light emerges (right side for light incident from left), the radius is positive. If it's on the side where light is incident (left side), the radius is negative.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires the use of the Lens Maker's Formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its two surfaces. It is essential to use the correct sign convention for the radii of curvature.

Step 2: Key Formula or Approach:
The Lens Maker's Formula is given by: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where: - \(f\) is the focal length of the lens.
- \(\mu\) is the refractive index of the lens material with respect to the surrounding medium.
- \(R_1\) is the radius of curvature of the first surface (where light enters).
- \(R_2\) is the radius of curvature of the second surface.

Step 3: Detailed Explanation:
Let's apply the sign convention for a double convex lens. We assume light travels from left to right.
1. For the first surface, the center of curvature is on the right side. Thus, its radius of curvature is positive: \(R_1 = +R\).
2. For the second surface, the center of curvature is on the left side. Thus, its radius of curvature is negative: \(R_2 = -R\).
The given values are: - Refractive index, \(\mu = 1.5\).
- Radius of curvature for both surfaces has magnitude \(R\).
Now, substitute these values into the Lens Maker's Formula: \[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{+R} - \frac{1}{-R} \right) \] \[ \frac{1}{f} = (0.5) \left( \frac{1}{R} + \frac{1}{R} \right) \] \[ \frac{1}{f} = (0.5) \left( \frac{2}{R} \right) \] \[ \frac{1}{f} = \frac{1}{2} \times \frac{2}{R} \] \[ \frac{1}{f} = \frac{1}{R} \] Taking the reciprocal of both sides, we get: \[ f = R \]

Step 4: Final Answer:
The focal length of the double convex lens is equal to its radius of curvature, \(f = R\).