Chapter 36, Problem 75AP

(a)

To determine

The maximum angular magnification produced in a telescope.

Answer to Problem 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 $1.50$.

(b)

To determine

The maximum overall magnification produced in a microscope.

Answer to Problem 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 $1.90$.