91Ó°ÊÓ

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.