Chapter 36, Problem 36.75AP

Andy decides to use an old pair of eyeglasses to make some optical instruments. He knows that the near point in his left eye is 50.0 cm and the near point in his right eve is 100 cm. (a) What is the maximum angular magnification he can produce in a telescope? (b) If he places the lenses 10.0 cm apart, what is the maximum overall magnification he can produce in a microscope? Hint: Go back to basics and use the thin lens equation to solve part (b).

To determine

The maximum angular magnification produced in a telescope.

Answer to Problem 36.75AP

The maximum angular magnification produced in a telescope is $1.50$.

Explanation of Solution

Given info: The near point in the left eye is $50.0\text{cm}$ and the near point in the right eye is $100\text{cm}$.

Write the thin lens equation.

$\begin{array}{c}\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\\ f=\frac{pq}{p+q}\end{array}$ (1)

Here,

$p$ is the object distance.

$q$ is the image distance.

$f$ is the focal length of the lens.

When the lens for left eye is objective lens.

Write the equation for focal length of the left lens.

$f_{\text{l}}=\frac{p_{\text{l}}{q}_{\text{l}}}{p_{\text{l}}+q_{\text{l}}}$

Here,

$p_{\text{l}}$ is the object distance and is equal to $25.0\text{cm}$.

$q_{\text{l}}$ is the image distance which is the near point in the left eye $-50.0\text{cm}$.

Substitute $25.0\text{cm}$ for $p_{\text{l}}$ and $-50.0\text{cm}$ for $q_{\text{l}}$ in the above equation to get the focal length of the left lens.

$\begin{array}{c}{f}_{\text{l}}=\frac{\left(25.0\text{cm}\right)\left(-50.0\text{cm}\right)}{\left(25.0\text{cm}\right)+\left(-50.0\text{cm}\right)}\\ =\frac{-1250{\text{cm}}^2}{-25\text{cm}}\\ =50\text{cm}\end{array}$

Write the equation for focal length of the right lens.

$f_{\text{r}}=\frac{p_{\text{r}}{q}_{\text{r}}}{p_{\text{r}}+q_{\text{r}}}$

Here,

$p_{\text{r}}$ is the object distance and is equal to $25.0\text{cm}$.

$q_{\text{r}}$ is the image distance which is the near point in the right eye $-100\text{cm}$.

Substitute $25.0\text{cm}$ for $p_{\text{r}}$ and $-100\text{cm}$ for $q_{\text{r}}$ in the above equation to get the focal length of the right lens.

$\begin{array}{c}{f}_{\text{l}}=\frac{\left(25.0\text{cm}\right)\left(-100\text{cm}\right)}{\left(25.0\text{cm}\right)+\left(-100\text{cm}\right)}\\ =\frac{-2500{\text{cm}}^2}{-75\text{cm}}\\ =33.3\text{cm}\end{array}$

Write the equation for angular magnification of a telescope when the lens for left eye is objective lens.

$m=\frac{f_{\text{o}}}{f_{\text{e}}}$

Here,

$f_{\text{o}}$ is the focal length of the objective lenses.

$f_{\text{e}}$ is the focal length of the eyepiece lenses.

Substitute $50\text{cm}$ for $f_{\text{o}}$ and $33.3\text{cm}$ for $f_{\text{e}}$ in the above equation to get the angular magnification of a telescope.

$\begin{array}{c}m=\frac{50\text{cm}}{33.3\text{cm}}\\ =1.50\end{array}$

Conclusion:

Therefore, the maximum angular magnification produced in a telescope is $1.50$.

To determine

The maximum overall magnification produced in a microscope.

Answer to Problem 36.75AP

The maximum overall magnification produced in a microscope is $1.90$.

Explanation of Solution

Given info: The near point in the left eye is $50.0\text{cm}$, the near point in the right eye is $100\text{cm}$ and the distance of lenses is $10.0\text{cm}$.

When the lens for right eye is eyepiece lens.

Write the thin lens equation to get the object distance for the eyepiece.

$\begin{array}{c}\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\\ p_{\text{e}}=\frac{q_{\text{e}}{f}_{\text{e}}}{q_{\text{e}}-f_{\text{e}}}\end{array}$

Here,

$q_{\text{e}}$ is the normal near point and is equal to $-25.0\text{cm}$.

$f_{\text{e}}$ is the focal length of the eyepiece lenses.

Substitute $25.0\text{cm}$ for $q_{\text{e}}$ and $33.3\text{cm}$ for $f_{\text{e}}$ in the above equation to get the object distance of the eyepiece lens.

$\begin{array}{c}{p}_{\text{e}}=\frac{\left(-25.0\text{cm}\right)\left(33.3\text{cm}\right)}{\left(-25.0\text{cm}\right)-\left(33.3\text{cm}\right)}\\ =\frac{-832.5{\text{cm}}^2}{-58.3\text{cm}}\\ \text{=14}\text{.3}\text{cm}\end{array}$

Write the equation for maximum magnification of the eyepiece.

$m_{\max }=1+\frac{25\text{cm}}{f_{\text{e}}}$

Substitute $33.3\text{cm}$ for $f_{\text{e}}$ in the above equation to get the maximum magnification of the eyepiece.

$\begin{array}{c}{m}_{\max }=1+\frac{25\text{cm}}{33.3\text{cm}}\\ =1.75\end{array}$

Calculate the image distance for the objective lens.

$q_{\text{l}}=L-p_{\text{e}}$

Substitute $10.0\text{cm}$ for $L$ and $14.3\text{cm}$ for $p_{\text{e}}$ in the above equation.

$\begin{array}{c}{q}_{\text{l}}=10.0\text{cm}-14.3\text{cm}\\ =-4.28\text{cm}\end{array}$

Write the thin lens equation to get the object distance for the objective lens.

$\begin{array}{c}\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\\ p_{\text{l}}=\frac{q_{\text{l}}{f}_{\text{l}}}{q_{\text{l}}-f_{\text{l}}}\end{array}$

Substitute $-4.28\text{cm}$ for $q_{\text{l}}$ and $50.0\text{cm}$ for $f_{\text{l}}$ in the above equation to get the object distance of the objective lens.

$\begin{array}{c}{p}_{\text{l}}=\frac{\left(-4.28\text{cm}\right)\left(50.0\text{cm}\right)}{\left(-4.28\text{cm}\right)-\left(50.0\text{cm}\right)}\\ =\frac{-214{\text{cm}}^2}{-54.28\text{cm}}\\ =\text{3}\text{.95}\text{cm}\end{array}$

Write the equation for magnification for objective lens.

$M=-\frac{q_{\text{l}}}{p_{\text{l}}}$

Substitute $-4.28\text{cm}$ for $q_{\text{l}}$ and $3.95\text{cm}$ for $p_{\text{l}}$ in the above equation.

$\begin{array}{c}M=-\frac{\left(-4.28\text{cm}\right)}{\left(3.95\text{cm}\right)}\\ =1.08\end{array}$

Write the equation for overall magnification.

$m=M{m}_{\max }$

Substitute $1.08$ for $M$ and $1.75$ for $m_{\max }$ in the above equation to get the maximum overall magnification produced in a microscope.

$\begin{array}{c}m=\left(1.08\right)\left(1.75\right)\\ =1.90\end{array}$

Conclusion:

Therefore, the maximum overall magnification produced in a microscope is $1.90$.