Question

An equiconcave lens of crown glass has to be formed. What should be the radii of the surfaces of the lens so that the power of the lens is \(-2.5\,\text{D}\)? The refractive index of crown glass is \(1.65\).

Hint:

Use \(P = (n-1)(1/R_1 - 1/R_2)\) with \(R_1=-R,\ R_2=+R\) so the bracket is \(-2/R\); solve for \(R\).

Answer 1

Step 1: Convert power to focal length.
Focal length \(f = \dfrac{1}{P} = \dfrac{1}{-2.5} = -0.4\ \text{m} = -40\ \text{cm}\). The negative sign confirms a diverging (concave) lens.

Step 2: Write the lens maker's formula in \(f\) form.
\(\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\).

Step 3: Use equal radii.
For an equiconcave lens \(R_1 = -R\) and \(R_2 = +R\), giving \(\dfrac{1}{R_1} - \dfrac{1}{R_2} = -\dfrac{2}{R}\).

Step 4: Substitute numbers in centimetres.
\(\dfrac{1}{-40} = (1.65-1)\left(-\dfrac{2}{R}\right) = 0.65\left(-\dfrac{2}{R}\right) = -\dfrac{1.30}{R}\).

Step 5: Solve.
\(-\dfrac{1}{40} = -\dfrac{1.30}{R}\ \Rightarrow\ R = 1.30\times 40 = 52\ \text{cm}\).
\[\boxed{R = 52\ \text{cm}\ (0.52\ \text{m})}\]