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

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

Every portion of a lens forms a complete image of the object because rays originating from every point on the object pass through every non-opaque part of the lens.

• Covering the lower half of the lens does not truncate or cut off any portion of the image; the entire object is still fully imaged. Hence (B) is incorrect.
• However, the effective aperture area admitting light from the object is halved. The total radiant flux converging to form the image is reduced by half, which means the brightness or intensity of the image is reduced ($I \propto A$). Hence (C) is correct.