Chapter 38, Problem 38SP

A convex-concave lens has faces of radii 3.0 and 4.0 cm, respectively, and is made of glass of refractive index 1.6. Determine (a) its focal length and (b) the linear magnification of the image when the object is 28 cm from the lens.

To determine

The focal length of a convex-concave lens having faces of radii $\text{3}\text{.0 cm}$ and $\text{4}\text{.0 cm}$, respectively, and the refractive index of the glass is $1.6$.

Answer to Problem 38SP

Solution:

$\text{+20 cm}$

Explanation of Solution

Given data:

The radius of the convex surface is $\text{3}\text{.0 cm}$.

The radius of the concave surface is $\text{4}\text{.0 cm}$.

The refractive index of the material is $1.6$.

Formula used:

Write the Lens maker’s equation for a spherical lens.

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

Here, $f$ is the focal length of the lens, $n$ is the refractive index of the lens material and $R_1$ is the radius of curvature of one spherical side, and $R_2$ is the radius of curvature of other spherical side. And, radius of curvature is positive if the center of curvature lies to the right of the surface, and negative when its center of curvature lies to the left of the surface.

Explanation:

Consider the Lens maker’s equation for a spherical lens.

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

Substitute $1.6$ for $n$, $\text{3}\text{.0 cm}$ for $R_1$, and $\text{4}\text{.0 cm}$ for $R_2$ since the radius of curvature of concave surface will lie right to the surface of lens.

$\begin{array}{c}\frac{1}{f}=\left(1.6-1\right)\left[\frac{1}{3.0\text{ cm}}-\frac{1}{4.0\text{ cm}}\right]\\ =0.6\left(\frac{4-3}{\left(3\right)\left(4\right)}\right)\\ =\frac{1}{20}\end{array}$

Solve for $f$.

$f=\text{ +20 cm}$

Positive sign indicates that the effect of both surface is convex.

Conclusion:

The focal length of a convex-concave lens is $\text{+20 cm}$.

To determine

The linear magnification of the image when the object is $\text{28 cm}$ from the lens.

Answer to Problem 38SP

Solution:

$-\text{2}\text{.5:1}$

Explanation of Solution

Given data:

The object distance is $\text{28 cm}$.

From the previous part, the focal length of a convex-concave lens is $\text{+20 cm}$.

Formula used:

Write the thin lens equation.

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Here, $f$ is the focal length of the lens, $s_o$ is the object distance from the lens, and $s_i$ is the image distance from the lens.

Write one expression for the magnification of an image.

$M_T=-\frac{s_i}{s_o}$

Here, $M_T$ is the magnification of the image, $s_i$ is the image distance from the lens, and $s_o$ is the object distance from the lens.

Explanation:

Consider the thin lens equation.

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Rearrange the above equation.

$\frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o}$

Substitute $\text{20 cm}$ for $f$ and $\text{28 cm}$ for $s_o$

$\frac{1}{s_i}=\frac{1}{\text{20 cm}}-\frac{1}{\text{28 cm}}$

Solve for $s_i$.

$s_i\text{ = 70 cm}$

Write one expression for the magnification of an image.

$M_T=-\frac{s_i}{s_o}$

Substitute $\text{70 cm}$ for $s_i$ and $\text{28 cm}$ for $s_o$

$\begin{array}{c}{M}_T=-\frac{\text{70 cm}}{\text{28 cm}}\\ =-\text{2}\text{.5}\end{array}$

Negative sign indicates that the image is real and inverted.

Conclusion:

Therefore, the linear magnification of the image for the convex-concave lens is $\text{2}\text{.5:1}$.