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: Lens maker's formula.
The power of a thin lens is \(P = \dfrac{1}{f}\), and
\[ P = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where \(n\) is the refractive index and \(R_1, R_2\) are the radii of the two surfaces (with sign convention).

Step 2: Apply the sign convention for an equiconcave lens.
For an equiconcave lens both surfaces have the same magnitude of radius \(R\). The first surface is concave toward the incoming light so \(R_1 = -R\), and the second surface \(R_2 = +R\). Then
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{-R} - \frac{1}{R} = -\frac{2}{R} \]

Step 3: Substitute the given values.
Here \(n = 1.65\) so \(n - 1 = 0.65\), and \(P = -2.5\,\text{D}\):
\[ -2.5 = 0.65\times\left(-\frac{2}{R}\right) \]

Step 4: Solve for \(R\).
\[ -2.5 = -\frac{1.30}{R} \]
\[ R = \frac{1.30}{2.5} = 0.52\ \text{m} \]

Step 5: Final answer.
\[ R = 0.52\ \text{m} = 52\ \text{cm} \]
Each surface of the equiconcave lens must have a radius of curvature of magnitude 52 cm.
\[\boxed{R = 52\ \text{cm}}\]