Chapter 32, Problem 82P

(a)

To determine

The focal length of eyepiece.

Explanation of Solution

Given:

Magnifying power of microscope is $600$ .

The angular magnification of eyepiece is $15.0$ .

The distance of objective lens from eyepiece is $22.0\text{cm}$ .

Formula used:

Write expression for angular magnification of eyepiece.

  $M_e=\frac{x}{f_e}$  ........ (1)

Here, $M_e$ is angular magnification of eyepiece, $x$ is far point of eye and $f_e$ is focal length of eyepiece.

Calculation:

Substitute $25\text{cm}$ for $x$ and $15$ for $M_e$ in equation (1) and solve for $f_e$ .

  $\begin{array}{c}15=\frac{25}{f_e}\\ f_e=1.67\text{cm}\end{array}$

Conclusion:

Thus, the focal length of eyepiece is $1.67\text{cm}$ .

(b)

To determine

The location object so that it is in focus for normal eye.

Explanation of Solution

Given:

Magnifying power of microscope is $600$ .

The angular magnification of eyepiece is $15.0$ .

The distance of objective lens from eyepiece is $22.0\text{cm}$ .

Formula used:

Write expression for angular magnification of eyepiece.

  $M_e=\frac{x}{f_e}$  ........ (1)

Here, $M_e$ is angular magnification of eyepiece, $x$ is far point of eye and $f_e$ is focal length of eyepiece.

Write expression for image distance.

  $v=L-f_e$

Write expression for magnifying power of microscope.

  $M=m_o{m}_e$

Rearrange above expression for $m_o$ .

  $m_o=\frac{M}{m_e}$

Substitute $-\frac{v}{u}$ for $m_o$ in above expression.

  $-\frac{v}{u}=\frac{M}{m_e}$

Substitute $L-f_e$ for $v$ in above expression.

  $-\frac{L-f_e}{u}=\frac{M}{m_e}$

Rearrange above expression for $u$ .

  $u=\frac{m_e\left(f_e-L\right)}{M}$  ........ (2)

Calculation:

Substitute $25\text{cm}$ for $x$ and $15$ for $M_e$ in equation (1) and solve for $f_e$ .

  $\begin{array}{c}15=\frac{25}{f_e}\\ f_e=1.67\text{cm}\end{array}$

Substitute $15.0$ for $m_e$ , $1.67$ for $f_e$ , $22.0$ for $L$ and $-600$ for $M$ in equation (2).

  $\begin{array}{l}u=\frac{\left(15\right)\left(1.67-22\right)}{-600}\\ u=0.508\text{cm}\end{array}$

Conclusion:

Thus, the object is $0.508\text{cm}$ from the objective lens.

(c)

To determine

The focal length of objective lens.

Explanation of Solution

Given:

Magnifying power of microscope is $600$ .

The angular magnification of eyepiece is $15.0$ .

The distance of objective lens from eyepiece is $22.0\text{cm}$ .

Formula used:

Write expression for angular magnification of eyepiece.

  $M_e=\frac{x}{f_e}$  ........ (1)

Here, $M_e$ is angular magnification of eyepiece, $x$ is far point of eye and $f_e$ is focal length of eyepiece.

Write expression for image distance.

  $v=L-f_e$

Write expression for magnifying power of microscope.

  $M=m_o{m}_e$

Rearrange above expression for $m_o$ .

  $m_o=\frac{M}{m_e}$

Substitute $-\frac{v}{u}$ for $m_o$ in above expression.

  $-\frac{v}{u}=\frac{M}{m_e}$

Substitute $L-f_e$ for $v$ in above expression.

  $-\frac{L-f_e}{u}=\frac{M}{m_e}$

Rearrange above expression for $u$ .

  $u=\frac{m_e\left(f_e-L\right)}{M}$  ........ (2)

Write expression for lens equation for objective lens.

  $f_o=\frac{vu}{v+u}$  ........ (3)

Calculation:

Substitute $25\text{cm}$ for $x$ and $15$ for $M_e$ in equation (1) and solve for $f_e$ .

  $\begin{array}{c}15=\frac{25}{f_e}\\ f_e=1.67\text{cm}\end{array}$

Substitute $15.0$ for $m_e$ , $1.67$ for $f_e$ , $22.0$ for $L$ and $-600$ for $M$ in equation (2).

  $\begin{array}{l}u=\frac{\left(15\right)\left(1.67-22\right)}{-600}\\ u=0.508\text{cm}\end{array}$

Substitute $0.508\text{cm}$ for $u$ , $22-1.67$ for $v$ in equation (3).

  $\begin{array}{l}{f}_o=\frac{\left(0.508\right)\left(22-1.67\right)}{\left(22-1.67\right)+\left(0.508\right)}\\ f_o=0.496\text{cm}\end{array}$

Conclusion:

Thus, the focal length of objective lens is $0.496\text{cm}$ .