From the thin lens formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Multiplying through by $v$:

$$1 - \frac{v}{u} = \frac{v}{f}$$

Since magnification $m = \frac{v}{u}$, this gives:

$$1 - m = \frac{v}{f} \implies m = 1 - \frac{v}{f}$$

Taking the magnitude/slope of the linear plot of $m$ versus $v$:

$$\left|\frac{dm}{dv}\right| = \frac{1}{f}$$

From the given graph:

• Vertical change (change in magnification, $\Delta m$) = $c$
• Horizontal change (change in image distance, $\Delta v$) = $b$

The magnifying power of an astronomical refracting telescope in normal adjustment is given by:

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

where:

• $f_o$ is the focal length of the objective lens
• $f_e$ is the focal length of the eyepiece lens

To increase the magnifying power $m$, one must increase the focal length of the objective lens ($f_o$) or decrease the focal length of the eyepiece ($f_e$). Hence, option 1 is the correct answer.

By the Lens Maker's formula:

$$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a plano-convex lens with convex surface facing light:

• $R_1 = +20\,\text{cm}$
• $R_2 = \infty$ (plane surface)
• $\mu = 1.5$

Substituting the values:

$$\frac{1}{f} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{\infty}\right) = 0.5 \times \frac{1}{20} = \frac{1}{40}\quad\implies\quad f = +40\,\text{cm}$$

For a convex mirror, the focal length is positive ($+f$).

• Object distance: $u = -f$
• Object height: $h_o = 1\,\text{m}$

Using the mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{f} = \frac{1}{f} \implies \frac{1}{v} = \frac{2}{f} \implies v = \frac{f}{2}$$

Linear magnification is:

$$m = -\frac{v}{u} = -\frac{f/2}{-f} = +\frac{1}{2} = 0.50$$

Height of the image:

$$h_i = m \times h_o = 0.50 \times 1\,\text{m} = 0.50\,\text{m}$$