The formula for refraction at a spherical surface separating medium 1 ($n_1 = 1$) and medium 2 ($n_2 = 4/3$) is:

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

According to Cartesian sign conventions:

• Object distance $u = -20\,\text{cm}$
• Radius of curvature $R = -10\,\text{cm}$ (center of curvature on the incident light side)
• Refractive indices: $n_1 = 1$ (air) and $n_2 = 4/3$
  Substituting these values:
  $$\frac{4/3}{v} - \frac{1}{-20} = \frac{4/3 - 1}{-10}$$
  $$\frac{4}{3v} + \frac{1}{20} = \frac{1/3}{-10} = -\frac{1}{30}$$
  $$\frac{4}{3v} = -\frac{1}{30} - \frac{1}{20} = \frac{-2 - 3}{60} = -\frac{5}{60} = -\frac{1}{12}$$
  $$\frac{1}{v} = -\frac{3}{4} \times \frac{1}{12} = -\frac{1}{16} \implies v = -16\,\text{cm}$$
  The negative sign indicates a virtual image formed $16\,\text{cm}$ to the left of $P$ (in air).

According to the Lens Maker's Formula, the focal length $f$ and optical power $P$ of a thin lens in air are given by:

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

When the radii of curvature $R_1$ and $R_2$ are held constant, the bracketed term is constant:

$$P \propto (\mu - 1)$$

From the thin lens formula:

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

Rearranging gives:

$$(v - f)(u + f) = f^2$$

This is the standard equation of a rectangular hyperbola shifted with asymptotes at $u = -f$ and $v = +f$. In magnitude terms for real images ($u$ vs $v$ with positive coordinate axes):

• When $u = f$, $v \rightarrow \infty$.
• When $u = 2f$, $v = 2f$.
• When $u \rightarrow \infty$, $v \rightarrow f$.
  This produces the asymptotic rectangular hyperbola shown in Option 2.

For an equilateral prism, the refracting angle is $A = 60^\circ$.
The prism formula is:

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Given $\mu = \sqrt{2}$ and $A = 60^\circ$:

$$\sqrt{2} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{0.5}$$ $$\sin\left(\frac{60^\circ + \delta_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:

$$\frac{60^\circ + \delta_m}{2} = 45^\circ \implies 60^\circ + \delta_m = 90^\circ \implies \delta_m = 30^\circ$$