For each thin lens shown in Fig. E34.37, calculate the location of the image of an object that is 1.80 cmto the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.

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

Substituting the given values in the equation

$\begin{array}{r}\frac{1}{f}=(1.5-1.0)\left(\frac{1}{10.0\mathrm{cm}}-\frac{1}{-15.0\mathrm{cm}}\right)\\ \frac{1}{f}=\frac{1}{12.0\mathrm{cm}}\\ \Rightarrow f=+12.0\mathrm{cm}\end{array}$

Use the relationship between the object’s distance s and the image distance s' for the spherical mirrors

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

Substitute the values to find the location

$\begin{array}{r}{s}^{\mathrm{\prime }}=\frac{sf}{s-f}\\ s^{\mathrm{\prime }}=\frac{\left(18.0\mathrm{cm}\right)\left(12.0\mathrm{cm}\right)}{18.0\mathrm{cm}-12.0\mathrm{cm}}\\ \Rightarrow s^{\mathrm{\prime }}=+36.0\mathrm{cm}\end{array}$

The positive sign signifies that the image of the object is to the right.

The radii of curvatures are $R_1=10cm$and $R_2=\infty$as the right side is flat

Use the lensmaker’s equation

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

Substitute the given values

$\begin{array}{l}\frac{1}{f}=\left(1.5-1.0\right)\left(\frac{1}{10.0\mathrm{cm}}-\frac{1}{\infty }\right)\\ \frac{1}{f}=\frac{1}{20.0\mathrm{cm}}\\ \Rightarrow f=20.0\mathrm{cm}\end{array}$

Use the relationship between the object’s distance s and the image distance s' for the spherical mirrors

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

Substitute the values to find the location

$\begin{array}{l}s\text{'}=\frac{(18.0\mathrm{cm})(20.0\mathrm{cm})}{18.0\mathrm{cm}-20.0\mathrm{cm}}\\ s\text{'}=-180cm\end{array}$

The negative signs signifies that the image is to the left.

The radii of the curvatures in this case are $R_1=10cmand{R}_2=15cm$, also the lens is a diverging lens, hence its focal length is negative.

Use the lensmaker’s equation

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

Substitute the given values

$\begin{array}{l}\frac{1}{f}=-\left(1.5-1.0\right)\left(\frac{1}{10.0\mathrm{cm}}-\frac{1}{-15.0\mathrm{cm}}\right)\\ \frac{1}{f}=\frac{1}{-12.0\mathrm{cm}}\\ \Rightarrow f=-12.0\mathrm{cm}\end{array}$

Use the relationship between the object’s distance s and the image distance s' for the spherical mirrors

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

Substitute the values to find the location

role="math" localid="1668163622815" $\begin{array}{l}s\text{'}=\frac{(18.0\mathrm{cm})(x-12.0\mathrm{cm})}{18.0\mathrm{cm}+12.0\mathrm{cm}}\\ \Rightarrow s\text{'}=-7.20\mathrm{cm}\end{array}$

The negative on the term signifies that the image is to the left.

Both the radii in this case are negative, thus, the focal length will also be negative

Use the lensmaker’s equation

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

Substitute the given values

$$\begin{gathered} \begin{array}{l}\frac{1}{f}=(1.5-1.0)\left(\frac{1}{-10.0\mathrm{cm}}-\frac{1}{-15.0\mathrm{cm}}\right)\\ \frac{1}{f}=\frac{1}{-60.0cm \\ }\\ f=-60.0cm\end{array} \end{gathered}$$

Use the relationship between the object’s distance s and the image distance s' for the spherical mirrors

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

Substitute the values to find the location

$\begin{array}{l}s\text{'}=\frac{(18.0\mathrm{cm})(-60.0\mathrm{cm})}{18.0\mathrm{cm}+60.0\mathrm{cm}}\\ s\text{'}=-13.8\mathrm{cm}\end{array}$

The negative on the term signifies that the image is to the left.