Question

A lens is made of glass having an index of refraction of 1.5. One side of the lens is flat, and the other is convex with a radius of curvature of 20 cm(a) Find the focal length of the lens. (b) If an object is placed 40 cmin front of the lens, where is the image?

Short answer

1. The focal length of the lens is +40cm.
2. If an object is placed 40 cm in front of the lens, then the image will be at $\infty$.

Step-by-step solution

Listing the given quantities

$$\begin{gathered} \mathrm{n}=1.5 \\ {\mathrm{r}}_1=\infty \\ {\mathrm{r}}_2=-20\mathrm{cm} \\ \mathrm{p}=40\mathrm{cm} \end{gathered}$$

Understanding the concepts of lens maker equation

Using lens maker’s equation, we can find the focal length for the given lens. We can find the image distance by using the equation for the thin lens.

Formula:

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{2}}}\mathbf{)} \\ \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{p}} \end{gathered}$$

Calculations of the focal length of the lens

(a) The lens maker’s equation is given by

$$\begin{gathered} \frac{1}{\mathrm{f}}=(\mathrm{n}-1)(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}) \\ \frac{1}{\mathrm{f}}=(1.5-1)(\frac{1}{\infty }-\frac{1}{-20}) \\ \frac{1}{\mathrm{f}}=0.5\times 0.05 \\ \mathrm{f}=\frac{1}{0.025} \end{gathered}$$

$=+40\mathrm{cm}$

Calculations of the image distance 

(b)

We have, for thin lens,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{i}}+\frac{1}{\mathrm{p}}$

Rearranging the terms,

$$\begin{gathered} \mathrm{i}=\frac{1}{\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{p}}} \\ =\frac{1}{\frac{1}{40}-\frac{1}{40}} \\ =\frac{1}{0} \\ =\infty \end{gathered}$$

If an object is placed 40 cm in front of the lens, then the image will be at$\infty$.