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).
गोलीय अपवर्तक पृष्ठ के सूत्र के अनुसार:
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$चिह्न परिपाटी के अनुसार:
$u = -20\,\text{cm}$, $R = -10\,\text{cm}$, $n_1 = 1$, $n_2 = 4/3$
मान रखने पर:
ऋणात्मक चिह्न यह दर्शाता है कि आभासी प्रतिबिंब ध्रुव $P$ से $16\,\text{cm}$ बाईं ओर (वायु में) बनता है।