Chapter 25, Problem 39P

(a)

To determine

The maximum angular magnification the telescope can produce.

Answer to Problem 39P

The maximum angular magnification the telescope can produce is $1.50$.

Explanation of Solution

The focal length of the lens for the left eye is,

$f_L=\frac{p_L{q}_L}{p_L+q_L}$

• • $p_L$ is the object distance
  • $q_L$ is the image distance

Substitute $25.0\text{cm}$ for $p_L$ and $-50.0\text{cm}$ for $q_L$.

$\begin{array}{c}{f}_L=\frac{\left(25.0\text{cm}\right)\left(-50.0\text{cm}\right)}{25.0\text{cm}-50.0\text{cm}}\\ =50.0\text{cm}\end{array}$

The focal length of the lens for the right eye is,

$f_R=\frac{p_R{q}_R}{p_R+q_R}$

Substitute $25.0\text{cm}$ for $p_R$ and $-100.0\text{cm}$ for $q_R$.

$\begin{array}{c}{f}_R=\frac{\left(25.0\text{cm}\right)\left(-100.0\text{cm}\right)}{25.0\text{cm}-100.0\text{cm}}\\ =33.3\text{cm}\end{array}$

Using the lens for the left eye as objective,

$m=\frac{f_o}{f_e}=\frac{f_L}{f_R}$

Substitute $50.0\text{cm}$ for $f_L$ and $33.3\text{cm}$ for $f_R$.

$\begin{array}{c}m=\frac{50.0\text{cm}}{33.3\text{cm}}\\ =1.50\end{array}$

Conclusion:

Thus, the maximum angular magnification the telescope can produce is $\underset{\_}{1.50}$.

(b)

To determine

The maximum overall magnification the telescope can produce, if the lenses are $10.0\text{cm}$ apart.

Answer to Problem 39P

The maximum overall magnification the telescope can produce, if the lenses are $10.0\text{cm}$ apart is $1.9$.

Explanation of Solution

Using the lens for the right eye as eyepiece, for maximum magnification, the final image must be formed at the normal near point.

The object distance is,

$p_e=\frac{q_e{f}_e}{q_e-f_e}$

Substitute $33.3\text{cm}$ for $f_e$ and $-25.0\text{cm}$ for $q_e$.

$\begin{array}{c}{p}_e=\frac{\left(-25.0\text{cm}\right)\left(33.3\text{cm}\right)}{-25.0\text{cm}-33.3\text{cm}}\\ =14.3\text{cm}\end{array}$

The maximum magnification by the eyepiece is,

$m_e=1+\frac{25.0\text{cm}}{f_e}$

Substitute $33.3\text{cm}$ for $f_e$.

$\begin{array}{c}{m}_e=1+\frac{25.0\text{cm}}{33.3\text{cm}}\\ =1.75\end{array}$

The image distance for the objective is,

$\begin{array}{c}{q}_1=L-p_e\\ =10.0\text{cm}-14.3\text{cm}\\ \text{=}-4.3\text{cm}\end{array}$

The object distance for the objective is,

$p_1=\frac{q_1{f}_1}{q_1-f_1}$

Substitute $50.0\text{cm}$ for $f_e$ and $-4.3\text{cm}$ for $q_e$.

$\begin{array}{c}{p}_1=\frac{\left(-4.3\text{cm}\right)\left(50.0\text{cm}\right)}{-4.3\text{cm}-50.0\text{cm}}\\ =4.0\text{cm}\end{array}$

The magnification by the objective is,

$\begin{array}{c}{M}_1=-\frac{q_1}{p_1}\\ =-\frac{-4.33\text{cm}}{4.0\text{cm}}\\ =1.1\end{array}$

The overall magnification is,

$m=M_1{m}_e$

Substitute $1.1$ for $M_1$ and $1.75$ for $m_e$.

$\begin{array}{c}m=\left(1.1\right)\left(1.75\right)\\ =1.9\end{array}$

Conclusion:

Thus, the maximum overall magnification the telescope can produce, if the lenses are $10.0\text{cm}$ apart is $\underset{\_}{1.9}$