Chapter 24, Problem 87QAP

To determine

The possible value of its focal length and sketch the cross-sectional sketch of the lens.

Answer to Problem 87QAP

  $R_1=12cm$ and $R_2=18cm$ ; $f=60mm$

  $R_1=-12cm$ And $R_2=-18cm$ ; $f=-60mm$

Explanation of Solution

Given:

  $n=2.16$

Radius of curvature

  $R_1=12cm$

  $R_2=18cm$$$

Formula used:

  $\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Calculation:

Index of refraction:

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

Possible focal lengths assuming both radius of curvature are positive:

  $R_1=12cm$ And $R_2=18cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(12\right)\left(18\right)}{\left(1.60-1\right)\left(18-12\right)}\\ f=60mm\end{array}$

  $R_1=18cm$ And $R_2=12cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(18\right)\left(12\right)}{\left(1.60-1\right)\left(12-18\right)}\\ f=-60mm\end{array}$

Possible focal length assuming both radius of curvature are negative:

  $R_1=-12cm$ And $R_2=-18cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(-12\right)\left(-18\right)}{\left(1.60-1\right)\left(-18\right)-\left(-12\right)}\\ f=-60mm\end{array}$

  $R_1=-18cm$ And

  $R_2=-12cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(-18\right)\left(-12\right)}{\left(1.60-1\right)\left(-12\right)-\left(-18\right)}\\ f=60mm\end{array}$

Possible focal length assuming both radius of curvature is negative and other are positive:

  $R_1=12cm$ and $R_2=-18cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(-12\right)\left(-18\right)}{\left(1.60-1\right)\left(-18\right)-\left(12\right)}\\ f=12mm\end{array}$

If $R_1=-18cm$ and $R_2=12cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(-18\right)\left(12\right)}{\left(1.60-1\right)\left(12\right)-\left(-18\right)}\\ f=-12mm\end{array}$

If $R_1=18cm$ and $R_2=-12cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(18\right)\left(-12\right)}{\left(1.60-1\right)\left(-12\right)-\left(18\right)}\\ f=12mm\end{array}$

If $R_1=-12cm$ and $R_2=18cm$

  $f=\frac{R_1{R}_2}{\left(n-1\right)\left(R_2-R_1\right)}$

  $\begin{array}{l}f=\frac{\left(-12\right)\left(18\right)}{\left(1.60-1\right)\left(18\right)-\left(-12\right)}\\ f=-12mm\end{array}$

Sketch: