Question

Consider a thin bi-convex lens of relative refractive index \(n = 1.5\). The radius of curvature of one surface of the lens is twice that of the other. The magnitude of larger radius of curvature in units of the focal length of the lens is ............. (Round off to 1 decimal place).

Hint:

For a bi-convex lens, always assign opposite signs to the two radii of curvature when using the lens maker's equation.

Answer 1

Correct Answer: 1.4

Step 1: Lens maker's formula.
\[ \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] Let \(R_2 = 2R_1 = 2R\). Since the lens is bi-convex, both surfaces bulge outwards. Thus, \(R_1 = +R,\ R_2 = -2R\).

Step 2: Substitute values.
\[ \frac{1}{f} = (1.5 - 1)\left(\frac{1}{R} - \frac{1}{-2R}\right) \] \[ \frac{1}{f} = 0.5\left(\frac{3}{2R}\right) = \frac{3}{4R} \] \[ R = \frac{3}{4}f \] Larger radius of curvature = \(2R = \frac{3}{2}f = 1.5f\).

Step 3: Express in magnitude.
\[ \text{Hence, } \frac{\text{Larger radius}}{f} = 1.5 \] Rounded to one decimal place: 1.5.