A surveyor on one bank of canal observed the image of the 4 inch and 17 ft marks on a vertical staff,which is partially immersed in the water and held against the bank directly opposite to him, coincides If the `17 ft` mark and the surveyor's eye are both `6 ft` above the water level ,estimate the width of the canal ,assuming that the reflective index of the water is `4//3` .Zero mark is at the bottom of the canal.

Explanation

To solve the problem, we need to estimate the width of the canal based on the observations made by the surveyor. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
The surveyor observes two marks on a vertical staff: one at 4 inches (which is \( \frac{1}{3} \) ft) and the other at 17 ft. Both the surveyor's eye and the 17 ft mark are 6 ft above the water level.

### Step 2: Convert Measurements
Convert the 4-inch mark to feet:
\[ 4 \text{ inches} = \frac{4}{12} \text{ ft} = \frac{1}{3} \text{ ft} \]

### Step 3: Determine the Height of the Marks Above the Bottom of the Canal
The height of the 17 ft mark above the bottom of the canal:
\[ \text{Height of 17 ft mark} = 6 \text{ ft (above water)} + 0 \text{ ft (water level)} = 6 \text{ ft} \]
The height of the 4-inch mark above the bottom of the canal:
\[ \text{Height of 4-inch mark} = 6 \text{ ft} - \frac{1}{3} \text{ ft} = 6 - \frac{1}{3} = \frac{18}{3} - \frac{1}{3} = \frac{17}{3} \text{ ft} \]

### Step 4: Use Snell's Law
The refractive index of water is given as \( \mu = \frac{4}{3} \). According to Snell's law:
\[ \frac{\sin(\beta)}{\sin(\alpha)} = \frac{n_2}{n_1} = \frac{4/3}{1} = \frac{4}{3} \]
Where \( \beta \) is the angle of refraction (in air) and \( \alpha \) is the angle of incidence (in water).

### Step 5: Set Up the Geometry
Let \( d \) be the width of the canal. The height of the 17 ft mark above the water level is 6 ft, and the height of the 4-inch mark is \( \frac{17}{3} \) ft.

### Step 6: Calculate the Distances
Using the right triangle formed by the surveyor's line of sight to the marks:
- For the 17 ft mark:
\[ \sin(\alpha) = \frac{d/2}{\sqrt{(d/2)^2 + (32/3)^2}} \]
- For the 4-inch mark:
\[ \sin(\beta) = \frac{d/2}{\sqrt{(d/2)^2 + 6^2}} \]

### Step 7: Apply Snell's Law
Substituting the expressions for \( \sin(\alpha) \) and \( \sin(\beta) \) into Snell's law gives:
\[ \frac{d/2}{\sqrt{(d/2)^2 + (32/3)^2}} = \frac{4}{3} \cdot \frac{d/2}{\sqrt{(d/2)^2 + 6^2}} \]

### Step 8: Cross Multiply and Simplify
Cross-multiplying and simplifying leads to:
\[ \sqrt{(d/2)^2 + (32/3)^2} = \frac{3}{4} \sqrt{(d/2)^2 + 6^2} \]

### Step 9: Square Both Sides
Squaring both sides and simplifying:
\[ (d/2)^2 + \left(\frac{32}{3}\right)^2 = \frac{9}{16} \left((d/2)^2 + 36\right) \]

### Step 10: Solve for \( d \)
After simplifying the equation, we can solve for \( d \):
\[ 16(d/2)^2 + \frac{1024}{9} = 9(d/2)^2 + 81 \]
This leads to:
\[ 7(d/2)^2 = 81 - \frac{1024}{9} \]
Calculating the right side gives:
\[ d = 16 \text{ ft} \]

### Final Answer
The estimated width of the canal is:
\[ \boxed{16 \text{ ft}} \]