Chapter 32, Problem 45P

(a)

To determine

The position and size of the image.

Explanation of Solution

Given:

The height of the object is $3.00\text{cm}$ .

The distance of the object in front of a thin lens is $25.0\text{cm}$ .

The power of the lens is $10.0\text{D}$

Formula used:

Draw a ray diagram to show the position and size of the image.

  

Write the expression for thin lens equation.

  $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$   ....... (1)

Here, $u$ is the object distance, $v$ is the image distance and $f$ is the focal length of the lens.

Write the expression for the focal length in terms of power of a lens.

  $f=\frac{1}{P}$   ....... (2)

Here, $P$ is the power of the lens.

Write the expression for lateral magnification in terms of image height and object height.

  $m=\frac{y^{\prime }}{y}$

Here, $m$ is the lateral magnification, $y^{\prime }$ is the image height and $y$ is the object height.

Write the expression for lateral magnification in terms of image distance and object distance.

  $m=-\frac{v}{u}$   ....... (3)

Combine the above two equations.

  $\begin{array}{l}\frac{y^{\prime }}{y}=-\frac{v}{u}\\ y^{\prime }=-\frac{v}{u}y\end{array}$   ....... (4)

Calculation:

Substitute $10.0\text{D}$ for $P$ in equation (2).

  $\begin{array}{l}f=\frac{1}{10.0\text{D}}\\ f=0.100\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ f=10.0\text{cm}\end{array}$

Rewrite equation (1) to calculate the image distance.

  $v=\frac{fu}{u-f}$

Substitute $10.0\text{cm}$ for $f$ and $25.0\text{cm}$ for $u$ in the above equation.

  $\begin{array}{l}v=\frac{\left(10.0\text{cm}\right)\left(25.0\text{cm}\right)}{25.0\text{cm}-10.0\text{cm}}\\ v\approx 16.67\text{cm}\end{array}$

Substitute $16.7\text{cm}$ for $v$ and $25.0\text{cm}$ for $u$ in equation (3).

  $\begin{array}{l}m=-\frac{16.7\text{cm}}{25.0\text{cm}}\\ m\approx -0.67\end{array}$

Substitute $16.7\text{cm}$ for $v$ , $3.00\text{cm}$ for $y$ and $25.0\text{cm}$ for $u$ in equation (4).

  $\begin{array}{l}I=-\frac{16.7\text{cm}}{25.0\text{cm}}\left(3.00\text{cm}\right)\\ I\approx -2.00\text{cm}\end{array}$

Conclusion:

Thus, the image is real and inverted. As the image distance, $v>0$ so the image is real and the lateral magnification, $m<0$ so the image is inverted and diminished.

(b)

To determine

The position and size of the image.

Explanation of Solution

Given:

The height of the object is $3.00\text{cm}$ .

The distance of the object in front of a thin lens is $20.0\text{cm}$ .

The power of the lens is $10.0\text{D}$

Formula used:

Draw a ray diagram to show the position and size of the image.

  

Write the expression for thin lens equation.

  $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Write the expression for the focal length in terms of power of a lens.

  $f=\frac{1}{P}$

Write the expression for lateral magnification in terms of image height and object height.

  $m=\frac{y^{\prime }}{y}$

Write the expression for lateral magnification in terms of image distance and object distance.

  $m=-\frac{v}{u}$

Combine the above two equations.

  $\begin{array}{l}\frac{y^{\prime }}{y}=-\frac{v}{u}\\ y^{\prime }=-\frac{v}{u}y\end{array}$

Calculation:

Substitute $10.0\text{D}$ for $P$ in equation (2).

  $\begin{array}{l}f=\frac{1}{10.0\text{D}}\\ f=0.100\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ f=10.0\text{cm}\end{array}$

Rewrite equation (1) to calculate the image distance.

  $v=\frac{fu}{u-f}$

Substitute $10.0\text{cm}$ for $f$ and $20.0\text{cm}$ for $u$ in the above equation.

  $\begin{array}{l}v=\frac{\left(10.0\text{cm}\right)\left(20.0\text{cm}\right)}{20.0\text{cm}-10.0\text{cm}}\\ v=20.0\text{cm}\end{array}$

Substitute $20.0\text{cm}$ for $v$ and $20.0\text{cm}$ for $u$ in equation (3).

  $\begin{array}{l}m=-\frac{20.0\text{cm}}{20.0\text{cm}}\\ m\approx -1.00\end{array}$

Substitute $20.0\text{cm}$ for $v$ , $3.00\text{cm}$ for $y$ and $20.0\text{cm}$ for $u$ in equation (4).

  $\begin{array}{l}{y}^{\prime }=-\frac{20.0\text{cm}}{20.0\text{cm}}\left(3.00\text{cm}\right)\\ y^{\prime }\approx -3.00\text{cm}\end{array}$

Conclusion:

Thus, the image is real and inverted. As the image distance, $v>0$ so the image is real and the lateral magnification, $m<0$ so the image is inverted and same size.

(c)

To determine

The position and size of the image.

Explanation of Solution

Given:

The height of the object is $3.00\text{cm}$ .

The distance of the object in front of a thin lens is $20.0\text{cm}$ .

The power of the lens is $-10.0\text{D}$

Formula used:

Draw a ray diagram to show the position and size of the image.

  

Write the expression for thin lens equation.

  $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Write the expression for the focal length in terms of power of a lens.

  $f=\frac{1}{P}$

Write the expression for lateral magnification in terms of image height and object height.

  $m=\frac{y^{\prime }}{y}$

Write the expression for lateral magnification in terms of image distance and object distance.

  $m=-\frac{v}{u}$

Combine the above two equations.

  $\begin{array}{l}\frac{y^{\prime }}{y}=-\frac{v}{u}\\ y^{\prime }=-\frac{v}{u}y\end{array}$

Calculation:

Substitute $\left(-10.0\text{D}\right)$ for $P$ in equation (2).

  $\begin{array}{l}f=\frac{1}{-10.0\text{D}}\\ f=-0.100\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ f=-10.0\text{cm}\end{array}$

Rewrite equation (1) to calculate the image distance.

  $v=\frac{fu}{u-f}$

Substitute $\left(-10.0\text{cm}\right)$ for $f$ and $20.0\text{cm}$ for $u$ in the above equation.

  $\begin{array}{l}v=\frac{\left(-10.0\text{cm}\right)\left(20.0\text{cm}\right)}{20.0\text{cm}-\left(-10.0\text{cm}\right)}\\ v\approx -6.67\text{cm}\end{array}$

Substitute $\left(-6.67\text{cm}\right)$ for $v$ and $20.0\text{cm}$ for $u$ in equation (3).

  $\begin{array}{l}m=-\frac{-6.67\text{cm}}{20.0\text{cm}}\\ m\approx 0.33\end{array}$

Substitute $\left(-6.67\text{cm}\right)$ for $v$ , $3.00\text{cm}$ for $y$ and $20.0\text{cm}$ for $u$ in equation (4).

  $\begin{array}{l}{y}^{\prime }=-\frac{\left(-6.67\text{cm}\right)}{20.0\text{cm}}\left(3.00\text{cm}\right)\\ y^{\prime }\approx 1.00\text{cm}\end{array}$

Conclusion:

Thus, the image is virtual and inverted. As the image distance, $v<0$ so the image is virtual and one third of the size of object.