Chapter 38, Problem 31SP

Determine the nature, position, and transverse magnification of the image formed by a thin converging lens of focal length +100 cm when the object distance from the lens is (a) 150 cm, (b) 75.0 cm.

To determine

The nature, position, and transverse magnification of an image formed by a thin converging lens of focal length $\text{100 cm}$ if the object distance is $\text{150 cm}$.

Answer to Problem 31SP

Solution:

Inverted, real image, $\text{300 cm}$ beyond the lens, and magnified by $2:1$.

Explanation of Solution

Given data:

The object distance is $\text{150 cm}$.

The focal length of the lens is $\text{100 cm}$.

Formula used:

Write the thin lens equation.

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

Rearrange the above equation.

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

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

Write the 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 distance of the image from the lens, and $s_o$ is the distance of the object from the lens.

Explanation:

Recall the thin lens equation.

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

Substitute $\text{100 cm}$ for $f$ and $\text{150 cm}$ for $s_o$

$\frac{1}{s_i}=\frac{1}{100\text{ cm}}-\frac{1}{150\text{ cm}}$

Solve for $s_i$.

$s_i=\text{300 cm}$

Write the expression for the magnification of an image.

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

Substitute $\text{300 cm}$ for $s_i$ and $\text{150 cm}$ for $s_o$

$\begin{array}{c}{M}_T=-\frac{300\text{ cm}}{150\text{ cm}}\\ =-\text{2}\end{array}$

Here, negative sign shows that the image is inverted. Therefore, the ration of the object’s magnification will be $2:1$.

Conclusion:

Therefore, the image shall be formed at a distance of $\text{300 cm}$ from the lens( on the left side) and shall be inverted, real, and magnified. The magnification of the image is $2:1$.

To determine

The nature, position and transverse magnification of an image formed by a thin converging lens of focal length $\text{100 cm}$ if the object distance is $\text{75}\text{.0 cm}$.

Answer to Problem 31SP

Solution:

$\text{300 cm}$ on the right side of the lens and erect, virtual, magnified by $4:1$.

Explanation of Solution

Given data:

The object distance is $\text{75}\text{.0 cm}$.

The focal length of the lens is $\text{100 cm}$.

Formula used:

Write the thin lens equation.

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

Rearrange the above equation.

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

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

Write the 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 distance the image from the lens, and $s_o$ is the distance of the object from the lens.

Explanation:

Recall the thin lens equation.

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

Substitute $\text{100 cm}$ for $f$ and $\text{75}\text{.0 cm}$ for $s_o$

$\frac{1}{s_i}=\frac{1}{100\text{ cm}}-\frac{1}{75.0\text{ cm}}$

Solve for $s_i$.

$s_i=-\text{300 cm}$

Write the expression for the magnification of an image.

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

Substitute $-\text{300 cm}$ for $s_i$ and $\text{75}\text{.0 cm}$ for $s_o$

$\begin{array}{l}{M}_T=-\frac{\left(-\text{300 cm}\right)}{75.0\text{ cm}}\\ \text{ = +4}\end{array}$

Here, positive sign shows that the image is erect. Therefore, the ration of the object’s magnification will be $4:1$.

Conclusion:

The image shall be formed at a distance of $-\text{300 cm}$ from the lens( on the right side) and shall be erect, virtual, and highly magnified. The magnification of the image is $4:1$.