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Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

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1. Given Data

  • Refractive Index (n): 1.55
  • Focal Length (f): 20 cm
  • Condition: Double-convex lens with equal radii (R1 = R, R2 = -R)

2. Formula

According to the Lens Maker's Formula:

1/f = (n - 1) × (1/R1 - 1/R2)

For a double-convex lens with equal curvature:

  • R1 is positive (+R)
  • R2 is negative (-R)

The formula simplifies to:

1/f = (n - 1) × (1/R - (-1/R))

1/f = (n - 1) × (2/R)

3. Calculation

Substituting the given values:

1/20 = (1.55 - 1) × (2/R)

1/20 = 0.55 × (2/R)

1/20 = 1.1 / R

Rearranging to find R:

R = 20 × 1.1

R = 22 cm

Required Radius of Curvature (R) = 22 cm

Conclusion: To achieve a focal length of 20 cm with this specific glass, both faces of the double-convex lens must be manufactured with a radius of curvature of 22 cm.

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