A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index -of water = 1-33.

Explanation

To solve the problem step by step, we will analyze the situation involving the cylindrical vessel filled with water and the object placed at the bottom.

### Step 1: Understand the Setup
We have a cylindrical vessel with a radius of 3 cm and a height of 4 cm filled with water. An object is placed at the center of the bottom of the vessel.

### Step 2: Identify the Rays
Two rays are considered:
1. A ray that leaves the object and exits the vessel through a corner.
2. A ray that travels directly along the axis of the vessel.

### Step 3: Determine the Real Depth
The real depth (d) of the object from the surface of the water is given as 4 cm (the height of the vessel).

### Step 4: Apply Snell's Law
Using Snell's law, we can relate the angles of incidence (I) and refraction (R) as follows:
\[ \frac{\sin I}{\sin R} = \frac{n_1}{n_2} \]
where \( n_1 \) is the refractive index of air (approximately 1) and \( n_2 \) is the refractive index of water (1.33).

### Step 5: Calculate Sine of Angles
To find the angle of incidence (I), we can use the geometry of the triangle formed by the radius and the height:
- The radius of the vessel is 3 cm, and the height is 4 cm.
- We can find the hypotenuse using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(3^2 + 4^2)} = \sqrt{9 + 16} = 5 \text{ cm} \]
- Therefore,
\[ \sin I = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} \]

### Step 6: Substitute into Snell's Law
Using Snell's law:
\[ \frac{3/5}{\sin R} = \frac{1}{1.33} \]
Thus,
\[ \sin R = \frac{3/5}{1.33} = \frac{3}{5 \times 1.33} = \frac{3}{6.65} \approx 0.451 \]

### Step 7: Calculate Apparent Depth
Using the relationship between the angles and the geometry of the situation, we can express the apparent depth (x):
\[ \tan R = \frac{3}{x} \]
From the triangle, we know:
\[ \tan R = \frac{\sin R}{\cos R} \]
Using the identity \( \sin^2 R + \cos^2 R = 1 \), we can find \( \cos R \) and hence calculate \( \tan R \).

### Step 8: Find x
We can equate:
\[ \tan R = \frac{3}{x} \quad \text{and} \quad \tan R = \frac{4}{3} \]
Thus,
\[ \frac{4}{3} = \frac{3}{x} \]
Cross-multiplying gives:
\[ 4x = 9 \implies x = \frac{9}{4} = 2.25 \text{ cm} \]

### Step 9: Calculate the Ratio of Real Depth to Apparent Depth
The ratio of real depth to apparent depth is:
\[ \text{Ratio} = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{4}{2.25} \approx 1.78 \]

### Final Answers
1. The apparent depth (x) is 2.25 cm.
2. The ratio of real depth to apparent depth is approximately 1.78.