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Section 47.8 Lensmaker's Equation

In Section 47.7 we stated that object distance \(p\text{,}\) image distance \(q\) and focal length \(f\) are related by

\begin{equation*} \frac{1}{p} + \frac{1}{q} = \frac{1}{f}. \end{equation*}

Focal length is a property of the lens and the media around the lens, being positive for convex lens and negative for concave lens.

It is possible to find an expression for the focal length in terms of the radii \(R_1\) and \(R_2\) of the two refracting surfaces of the lens, the refractive index \(n_0\) of the medium in which lens will be placed, and the refractive index \(n_l\) of the material of the lens (see Figure 47.8.1). If we ignore the thickness \(d\) of the lens, the formula for the inverse of the focal length is

\begin{equation} \frac{1}{f} = \left( \frac{n_l - n_0}{n_0} \right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right).\label{eq-focal-length-in-terms-of-radii}\tag{47.8.1} \end{equation}

This equation is called the lensmaker's equation. By sign convention of refraction from a curved surface, the radius is positive if the center falls on the right of the vertex and negative when it falls on the left.

\begin{align} \amp \text{ if center of sphere is on left then } R \lt 0.\tag{47.8.2}\\ \amp \text{ if center of sphere is on right then } R \gt 0.\tag{47.8.3} \end{align}

In Figure 47.8.1, \(R_1\gt 0\) and \(R_2\lt 0\text{.}\)

Figure 47.8.1.

If the lens is operating in air, then with \(n_0=1\text{,}\) this formula will simplify.

\begin{equation} \frac{1}{f} = \left( n_l - 1 \right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right).\label{eq-focal-length-in-terms-of-radii-in-air}\tag{47.8.4} \end{equation}

Even the formula with thickness \(d\) is not too complicated for lens in air.

\begin{equation} \frac{1}{f} = \left( n_l - 1 \right) \left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n_l-1) d}{ n_l R_1 R_2}\right].\label{eq-focal-length-in-terms-of-radii-in-air-with-thickness}\tag{47.8.5} \end{equation}

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. (47.8.4). 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)

With \(R_1=-22\text{ cm} = -R_2\) and \(n_0=1.33\) in Eq. (47.8.1), we get

\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, \(f = 66.7\text{ cm}\text{.}\)