Suppose you have been asked to make a bi-concave lens of focal length \(-20\text{ cm}\) in air.
(a) Find the radius of curvature of a symmetrically ground biconcave lens from a glass of refractive index \(1.55\) assuming you can ignore the thickness of the lens.
(b) If the lens is used underwater, what will be refractive index of the lens then?
Hint.
(a) Use thin lensmaker’s equation for air. (b) Use general thin lensmaker’s equation. Note that \(R_1 \lt 0\) and \(R_2 \gt 0\text{.}\)
Answer.
(a) \(22\text{ cm}\text{,}\) (b) \(66.7\text{ cm}\text{.}\)
Solution 1. (a)
In a biconcave lens, the radii will have the following signs: \(R_1 \lt 0\text{,}\) and \(R_2 \gt 0\text{.}\) You can convince yourself by drawing a picture and applying the sgn convention. Now, here \(|R_1|=|R_2|\text{.}\) Lets use a common symbol \(x \gt 0\) for this absolute value in Eq. (2.41). Therefore,
\begin{equation*}
R_1 = -x,\ \ R_2 = x.
\end{equation*}
I will work in \(\text{cm}\) units for the length.
\begin{equation*}
\frac{1}{-20} = \left( 1.55 - 1 \right) \left(-\frac{1}{x} - \frac{1}{x} \right) = -\frac{0.55\times 2}{x} = \frac{1.1}{x}.
\end{equation*}
Therefore
\begin{equation*}
x = 1.1\times 20 = 22\text{ cm}.
\end{equation*}
Solution 2. (b)
\begin{align*}
\frac{1}{f} \amp = \left( \frac{n_l - n_0}{n_0} \right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right)\\
\amp = \left( \frac{1.55 - 1.33}{1.33} \right) \times \frac{2}{-22}\\
\amp = -\frac{0.33}{22}
\end{align*}
Therefore, \(f = 66.7\text{ cm}\text{.}\)
