Apparent Depth Comparison in Glass Cylindrical Media with Different Curved Surfaces

To find the correct statements, we analyze the refraction of light at the top surface of each glass cylinder.

General Formula for Refraction at a Single Spherical Surface

The relation between object distance ($u$), image distance ($v$), refractive indices of the media, and radius of curvature ($R$) is given by: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$

Sign Convention:

• Light travels from point $X$ inside the glass ($n_1 = 1.5$) upwards into air ($n_2 = 1.0$). Therefore, the upward direction is positive.
• The pole (vertex) of each top surface is taken as the origin.
• The object $X$ is located below the top surface, so the object distance is: $u = -H = -30\text{ cm}$

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1. Cylinder I (Flat Top Surface)

For a flat surface, the radius of curvature $R \to \infty$. Substituting the values into the refraction formula: $\frac{1}{v_1} - \frac{1.5}{-30} = 0$ $\frac{1}{v_1} + \frac{1}{20} = 0 \implies v_1 = -20\text{ cm}$

Thus, the apparent depth $H_1$ is: $H_1 = |v_1| = 20\text{ cm}$

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2. Cylinder II (Convex Top Surface)

For the convex upper surface, the center of curvature lies inside the glass (below the vertex). Since this direction is opposite to the upward path of incident light: $R = -3\text{ m} = -300\text{ cm}$

Applying the refraction formula: $\frac{1}{v_2} - \frac{1.5}{-30} = \frac{1.0 - 1.5}{-300}$ $\frac{1}{v_2} + \frac{1}{20} = \frac{-0.5}{-300} = \frac{1}{600}$ $\frac{1}{v_2} = \frac{1}{600} - \frac{1}{20} = \frac{1 - 30}{600} = -\frac{29}{600}$ $v_2 = -\frac{600}{29}\text{ cm}$

Thus, the apparent depth $H_2$ is: $H_2 = |v_2| = \frac{600}{29}\text{ cm} \approx 20.69\text{ cm}$

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3. Cylinder III (Concave Top Surface)

For the concave upper surface, the center of curvature lies in air (above the vertex), which is in the direction of the incident light: $R = +3\text{ m} = +300\text{ cm}$

Applying the refraction formula: $\frac{1}{v_3} - \frac{1.5}{-30} = \frac{1.0 - 1.5}{+300}$ $\frac{1}{v_3} + \frac{1}{20} = \frac{-0.5}{300} = -\frac{1}{600}$ $\frac{1}{v_3} = -\frac{1}{600} - \frac{1}{20} = \frac{-1 - 30}{600} = -\frac{31}{600}$ $v_3 = -\frac{600}{31}\text{ cm}$

Thus, the apparent depth $H_3$ is: $H_3 = |v_3| = \frac{600}{31}\text{ cm} \approx 19.35\text{ cm}$

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Analysis of Options:

• Option A: Since $H_2 = 20.69\text{ cm}$ and $H_1 = 20\text{ cm}$, we have $H_2 > H_1$. (Correct)
• Option B: Since $H_3 = 19.35\text{ cm}$ and $H_1 = 20\text{ cm}$, we have $H_3 < H_1$. (Incorrect)
• Option C: Since $H_2 = 20.69\text{ cm}$ and $H_3 = 19.35\text{ cm}$, we have $H_2 > H_3$. (Correct)
• Option D: Calculating the difference $H_2 - H_1$: $H_2 - H_1 = \frac{600}{29} - 20 = \frac{20}{29}\text{ cm} \approx 0.69\text{ cm}$ Since $0.69\text{ cm} < 0.8\text{ cm}$, this inequality is false. (Incorrect)

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Correct Statements:

(A) and (C)