Step-by-step Solution

This problem involves a plano-convex lens and has two parts: an image formed by refraction through the lens, and a faint image formed by reflection from the curved surface. We will analyze these two phenomena separately to find the properties of the lens.

We will use the Cartesian sign convention where light is assumed to travel from left to right. The pole of the optical surface is the origin (0,0).

Part 1: Analysis of the Faint Image (Reflection)

The faint image is formed due to reflection from the convex surface of the lens. This surface acts as a convex mirror. A real object is placed in front of it.

1. Object and Image Distances:

   • The object is placed 30 cm away in front of the curved surface. So, the object distance is u = -30 cm.
   • For a real object, a convex mirror always forms a virtual, erect, and diminished image behind the mirror.
   • The faint image is observed at a distance of 10 cm from the lens. Since the image is virtual, it is formed on the side opposite to the reflecting surface, so the image distance is v = +10 cm.

2. Calculate the Radius of Curvature (R):

   • We use the mirror formula: $1/v + 1/u = 1/f_m = 2/R$
   • Substituting the values: $\frac{1}{10} + \frac{1}{-30} = \frac{2}{R}$ $\frac{3 - 1}{30} = \frac{2}{R}$ $\frac{2}{30} = \frac{2}{R}$ $\frac{1}{15} = \frac{2}{R} \implies R = 30 \text{ cm}$
   • The radius of curvature of the convex surface is 30 cm. Since the surface is convex to the incident light, R is positive.

3. Evaluate Options B and C:

   • Option B: States that the radius of curvature is 45 cm. Our calculated value is 30 cm. So, Option B is incorrect.
   • Option C: States that the faint image is erect and real. The image formed by a convex mirror for a real object is always erect and virtual. So, Option C is incorrect.

Part 2: Analysis of the Main Image (Refraction)

The main image is formed by refraction through the plano-convex lens. The image size is double the object size, which means the magnitude of magnification |m| = 2.

1. Two Possible Cases for Magnification:

   • Case 1: The image is real and inverted, so m = -2.
   • Case 2: The image is virtual and erect, so m = +2.

2. Analyze Case 1 (m = -2):

   • Object distance u = -30 cm.
   • Magnification m = v/u = -2. So, v = -2u = -2(-30) = +60 cm. The positive sign indicates a real image formed on the other side of the lens.
   • Using the lens formula 1/v - 1/u = 1/f: $\frac{1}{60} - \frac{1}{-30} = \frac{1}{f}$ $\frac{1}{60} + \frac{2}{60} = \frac{1}{f}$ $\frac{3}{60} = \frac{1}{f} \implies \frac{1}{f} = \frac{1}{20}$ $f = 20 \text{ cm}$
   • This matches Option D. Let's now find the refractive index n using the Lens Maker's formula.
   • For a plano-convex lens with light incident on the curved surface, $R_1 = R = +30$ cm and $R_2 = \infty$. $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ $\frac{1}{20} = (n - 1) \left( \frac{1}{30} - \frac{1}{\infty} \right)$ $\frac{1}{20} = \frac{n - 1}{30}$ $n - 1 = \frac{30}{20} = 1.5$ $n = 2.5$
   • This matches Option A.
   • So, this case is consistent with options A and D.

3. Analyze Case 2 (m = +2):

   • Object distance u = -30 cm.
   • Magnification m = v/u = +2. So, v = 2u = 2(-30) = -60 cm. The negative sign indicates a virtual image formed on the same side as the object.
   • Using the lens formula 1/v - 1/u = 1/f: $\frac{1}{-60} - \frac{1}{-30} = \frac{1}{f}$ $-\frac{1}{60} + \frac{2}{60} = \frac{1}{f}$ $\frac{1}{60} = \frac{1}{f} \implies f = 60 \text{ cm}$
   • This contradicts Option D (f = 20 cm).
   • Let's find the refractive index n for this case: $\frac{1}{60} = (n - 1) \left( \frac{1}{30} \right)$ $n - 1 = \frac{30}{60} = 0.5$ $n = 1.5$
   • This contradicts Option A (n = 2.5).
   • Therefore, this case is not consistent with the given options.

Conclusion

The only scenario that consistently matches the provided options is Case 1, where the image formed by refraction is real and inverted (m = -2). This leads to the following conclusions:

• The refractive index of the lens is n = 2.5. (Option A is true)
• The radius of curvature of the convex surface is R = 30 cm. (Option B is false)
• The faint image due to reflection is erect and virtual. (Option C is false)
• The focal length of the lens is f = 20 cm. (Option D is true)

Thus, the correct statements are A and D.