Chapter 32, Problem 88P

(a)

To determine

Image distance of the objective.

Explanation of Solution

Given:

The focal length of the objective is $100\text{cm}$ and the object distance is $30\text{m}$ .

Formula used:

A Galilean telescope is composed of two lenses: one is concave lens which serves as the eye piece while another one is convex lens which serves as the objective lens.

Write the expression for thin lens equation.

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

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

Rearrange above equation for the value of $v$ .

  $v=\frac{\left(f\right)}{\left(u-f\right)}\left(u\right)$   ....... (2)

Calculation:

Substitute $1\text{m}$ for $f$ and $30\text{m}$ for $u$ in equation (2).

  $\begin{array}{c}v=\frac{\left(1\text{m}\right)\left(30\text{m}\right)}{\left(30\text{m-1m}\right)}\\ =103.45\text{cm}\end{array}$

Conclusion:

Thus, the image of the objective is at the distance of $103.45\text{cm}$ .

(b)

To determine

Object distance for the eye piece so that the final image is at near point.

Explanation of Solution

Given:

The focal length of the eye piece is $-5\text{cm}$ and the final image is at near point.

Formula used:

A Galilean telescope is composed of two lenses: one is concave lens which serves as the eye piece while another one is convex lens which serves as the objective lens.

Write the expression for thin lens equation.

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

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

Rearrange above equation for the value of $u$ .

  $u=\frac{\left(f\right)}{\left(v-f\right)}\left(v\right)$   ....... (2)

Calculation:

Substitute $-5\text{cm}$ for $f$ and $-25\text{cm}$ for $v$ in equation (2).

  $\begin{array}{c}u=\frac{\left(-5\text{cm}\right)\left(-25\text{cm}\right)}{\left(-25\text{cm}-\left(-5\text{cm}\right)\right)}\\ =-6.25\text{cm}\end{array}$

Conclusion:

Object distance for the eye piece so that the final image is at near point is $-6.25\text{cm}$ .

(c)

To determine

Distance between the objective lens and the eye piece.

Explanation of Solution

Given:

The image of the objective is at the distance of $103\text{cm}$ , Object distance for the eye pieceis $-6.3\text{cm}$ .

Formula used:

Write the expression for the separation distance between the lenses.

  $D=v_{\text{objective}}+u_{\text{eyepiece}}$   ....... (1)

Here, $D$ is the separation distance between the lenses, $v_{\text{objective}}$ is the image distance for the objective lens and $u_{\text{eyepiece}}$ is the object distance for the eyepiece.

Calculation:

Substitute $103.45\text{cm}$ for $v_{\text{objective}}$ and $-6.25\text{cm}$ for $u_{\text{eyepiece}}$ in equation (1).

  $\begin{array}{c}D=103.45\text{cm}-6.25\text{cm}\\ =97.2\text{cm}\end{array}$

Conclusion:

Distance between the objective lens and the eye piece is $97.2\text{cm}$ .

(d)

To determine

Height of the final image, angular magnification.

Explanation of Solution

Given:

The object height is $1.5\text{m}$ .

Formula used:

Write the expression for magnification of telescope in terms of image size and object size.

  $M=\frac{I}{O}$   ....... (1)

Here, $M$ is the magnificationof telescope, $I$ is the image size and $O$ is the object size.

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

  $m=-\frac{v}{u}$   ....... (2)

Here, $m$ is the lateral magnification, $v$ is the image distance and $u$ is the object distance.

Write the expression for the magnification of telescope.

  $M=\left(m_{\text{objective}}\right)\left(m_{\text{eyepiece}}\right)$   ...... (3)

Here, $M$ is the magnification of telescope, $m_{\text{objective}}$ is the magnification of objective and $m_{\text{eyepiece}}$ is the magnification of eyepiece.

Calculation:

Substitute $m_{\text{objective}}$ for $m$ , $103.45\text{cm}$ for $v$ and $3000\text{cm}$ for $u$ in equation (2).

  $\begin{array}{c}{m}_{\text{objective}}=-\frac{\left(103.45\text{cm}\right)}{\left(3000\text{cm}\right)}\\ =-0.0345\end{array}$

Substitute $m_{\text{eyepiece}}$ for $m$ , $-25\text{cm}$ for $v$ and $\text{-6}\text{.25cm}$ for $u$ in equation (2).

  $\begin{array}{c}{m}_{\text{eyepiece}}=-\frac{\left(-25\text{cm}\right)}{\left(-6.25\text{cm}\right)}\\ =-4\end{array}$

Substitute $-4$ for $m_{\text{eyepiece}}$ and $-0.0345$ for $m_{\text{objective}}$ in equation (3).

  $\begin{array}{c}M=\left(-0.0345\right)\left(-4\right)\\ =0.138\end{array}$

Rearrange equation (1) for $I$ .

  $I=MO$

Substitute $0.138$ for $M$ , $150\text{cm}$ for $O$ in above equation.

  $\begin{array}{c}I=\left(0.138\right)\left(150\text{cm}\right)\\ =20.7\text{cm}\end{array}$

Write the expression for angular magnification of telescope.

  $M_{\theta }=\frac{{\theta }_{\text{e}}}{{\theta }_{\text{o}}}$   ...... (4)

Here, $M_{\theta }$ is the angular magnification of telescope, ${\theta }_{\text{e}}$ is the angle subtended byimage and ${\theta }_{\text{o}}$ is theangle subtended by object.

Write the expression for the angle subtended by eyepiece.

  ${\theta }_{\text{e}}={\tan }^{-1}\left(\frac{I\text{'}}{u_{eyepiece}}\right)$   ...... (5)

Here, ${\theta }_{\text{e}}$ is the angle subtended byimage, $I\text{'}$ is the height of final image in telescope and $u_{\text{eyepiece}}$ is the object distance for the eyepiece.

Write the expression for the angle subtended by object.

  ${\theta }_{\text{o}}=\left(\frac{I}{u_{\text{objective}}}\right)$   ...... (6)

Here, ${\theta }_{\text{o}}$ is the angle subtended by object, $I$ is the height of object and $u_{\text{objective}}$ is the object distance of objective.

Substitute ${\tan }^{-1}\left(\frac{I\text{'}}{u_{eyepiece}}\right)$ for ${\theta }_{\text{e}}$ and $\frac{I}{u_{\text{objective}}}$ for ${\theta }_{\text{o}}$ in equation (4).

  $\begin{array}{c}{M}_{\theta }=\frac{{\tan }^{-1}\left(\frac{I\text{'}}{u_{eyepiece}}\right)}{\left(\frac{I}{u_{\text{objective}}}\right)}\\ =\frac{u_{\text{objective}}}{I}{\tan }^{-1}\left(\frac{I\text{'}}{u_{eyepiece}}\right)\end{array}$   ...... (7)

Substitute $30\text{m}$ for $u_{\text{objective}}$ , $1.5\text{m}$ for $I$ , $20.7\text{cm}$ for $I\text{'}$ and $\text{-6}\text{.25cm}$ for $u_{\text{eyepiece}}$ in equation (7).

  $\begin{array}{c}{M}_{\theta }=\frac{30\text{m}}{1.5\text{m}}{\tan }^{-1}\left(\frac{20.7\text{cm}}{6.25\text{cm}}\right)\\ =26\end{array}$

Conclusion:

Height of the final image is $20.7\text{cm}$ and angular magnification is $26$ .