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 20 cm?
The radius of curvature required is 22 cm.
Explanation
This numerical problem applies the lens maker's formula from the textbook. For a double-convex lens with both faces having the same radius of curvature R, the first surface has R₁ = +R (convex towards incident light) and the second surface has R₂ = -R (following sign convention). The formula 1/f = (n-1)[1/R₁ - 1/R₂] simplifies to 1/f = 2(n-1)/R for such an equiconvex lens.
Solution Steps
Step 1: Given - Refractive index n = 1.55, Focal length f = 20 cm, Both faces have same radius R
Step 2: Apply lens maker's formula: 1/f = (n-1)[1/R₁ - 1/R₂]
Step 3: For double-convex lens with equal radii: R₁ = +R, R₂ = -R, so 1/f = (n-1)[1/R + 1/R] = 2(n-1)/R
Step 4: Substitute values: 1/20 = 2(1.55-1)/R = 2(0.55)/R = 1.1/R
Step 5: Solve for R: R = = 22 cm