The Galilean Telescope. Figure P34.100 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is $\mathbf{M}\mathbf{=}\mathbf{-}{\mathbf{f}}_{\mathbf{1}}\mathbf{/}{\mathbf{f}}_{\mathbf{2}}$. (b) A Galilean telescope is to be constructed with the same objective lens as in Exercise 34.61. What focal length should the eyepiece have if this telescope is to have the same magnitude of angular magnification as the one in Exercise 34.61? (c) Compare the lengths of the telescopes. Figure P34.100

The ratio of the distance of the image$\mathit{u}\mathbf{\text{'}}$to the ratio of the object $\mathit{u}$is known as magnification $\mathit{M}$.

$\mathbf{}\mathbf{M}\mathbf{=}\mathbf{-}\frac{\mathbf{u}\mathbf{\text{'}}}{\mathbf{u}}$

The ratio of real image $\mathbf{y}\mathbf{\text{'}}$and focal length ${\mathbf{f}}_{\mathbf{2}}$is the distance of the image $\mathit{u}\mathbf{\text{'}}$.

$\mathbf{u}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{y}\mathbf{\text{'}}}{{\mathbf{f}}_{\mathbf{2}}}$

The ratio of real image $\mathbf{y}\mathbf{\text{'}}$and focal length ${\mathbf{f}}_{\mathbf{1}}$is the distance of the object $\mathbf{u}$.

$\mathbf{u}\mathbf{=}\frac{\mathbf{y}\mathbf{\text{'}}}{{\mathbf{f}}_1}$

Given that,

$$\begin{gathered} M=6.33 \\ {f}_1=95\mathrm{cm} \end{gathered}$$

The magnification of the Telescope is:

$M=-\frac{u\text{'}}{u}$

Where, $u\text{'}$is $\frac{y\text{'}}{f_2}$and $u$is $\frac{y\text{'}}{f_2}$.

Substitute the $u\text{'}$and$u$ in magnification formula

$$\begin{gathered} M=-\frac{u\text{'}}{u} \\ =-\frac{\left(y\text{'}l{f}_2\right)}{\left(y\text{'}l{f}_1\right)} \\ =-\frac{f_1}{f_2} \end{gathered}$$

Hence, proved $M=-\frac{f_1}{f_2}$.

Given that,

$$\begin{gathered} M=6.33 \\ {f}_1=95\mathrm{cm} \end{gathered}$$

Substitute the values of $f_1$and $M$in$M=-\frac{f_1}{f_2}$

$$\begin{gathered} M=-\frac{f_1}{f_2} \\ 6.33=-\frac{95}{f_2} \\ {f}_2=-\frac{95}{6.33} \\ {f}_2=-15\mathrm{cm} \end{gathered}$$

Hence, the focal length of eyepiece should be -15cm.

The total focal length of telescope is:

$$\begin{gathered} f=f_1+f_2 \\ =95+\left(-15\right) \\ =82\mathrm{cm} \end{gathered}$$

Hence, the focal length of telescope is less than the Exercise 34.61.