A man of height 2 m stands on a straight road on a hot day. The vertical temperature in the air results in a variation of refractive index with height y as `mu =mu_(0)sqrt((1+ay))` where `mu_(0)` is the refractive index of air near the road and a=`2 xx 10^(-6)//m`. What is the actual length of the road, man is able to see

Options

A. 2000m
B. 390m
C. infinite distance
D. None of these

Explanation

To solve the problem step by step, we need to determine the actual length of the road that the man can see, taking into account the variation of the refractive index with height.

### Step 1: Understand the given parameters
- Height of the man (H) = 2 m
- Variation of refractive index with height (y):
\[ \mu = \mu_0 \sqrt{1 + ay} \]
where \( a = 2 \times 10^{-6} \, \text{m}^{-1} \).

### Step 2: Set up the relationship between the angles and the refractive indices
Using Snell's law, we have:
\[ \mu \sin \theta = \mu_0 \sin 90^\circ \]
Since \(\sin 90^\circ = 1\), this simplifies to:
\[ \mu \sin \theta = \mu_0 \]
Thus, we can express \(\sin \theta\) as:
\[ \sin \theta = \frac{\mu_0}{\mu} = \frac{1}{\sqrt{1 + ay}} \]

### Step 3: Relate the angles to the distance
Using the small angle approximation, we can relate the differential distances:
\[ \frac{dx}{dy} = \tan \theta \approx \sin \theta \]
This gives us:
\[ dx = \frac{dy}{\sqrt{1 + ay}} \]

### Step 4: Integrate to find the total distance
To find the total distance \(x\) that the man can see, we integrate from \(y = 0\) to \(y = 2\) (the height of the man):
\[ x = \int_0^2 \frac{dy}{\sqrt{1 + ay}} \]

### Step 5: Solve the integral
Substituting \(a = 2 \times 10^{-6}\):
\[ x = \int_0^2 \frac{dy}{\sqrt{1 + 2 \times 10^{-6} y}} \]

This integral can be solved using a substitution method. Let:
\[ u = 1 + 2 \times 10^{-6} y \implies du = 2 \times 10^{-6} dy \implies dy = \frac{du}{2 \times 10^{-6}} \]
When \(y = 0\), \(u = 1\) and when \(y = 2\), \(u = 1 + 4 \times 10^{-6}\).

Thus, the integral becomes:
\[ x = \int_1^{1 + 4 \times 10^{-6}} \frac{1}{\sqrt{u}} \cdot \frac{du}{2 \times 10^{-6}} = \frac{1}{2 \times 10^{-6}} \left[ 2\sqrt{u} \right]_1^{1 + 4 \times 10^{-6}} \]
Calculating this gives:
\[ x = \frac{1}{2 \times 10^{-6}} \left( 2\sqrt{1 + 4 \times 10^{-6}} - 2 \right) \]
Using the approximation \(\sqrt{1 + x} \approx 1 + \frac{x}{2}\) for small \(x\):
\[ \sqrt{1 + 4 \times 10^{-6}} \approx 1 + 2 \times 10^{-6} \]
Thus:
\[ x \approx \frac{1}{2 \times 10^{-6}} \left( 2(1 + 2 \times 10^{-6}) - 2 \right) = \frac{1}{2 \times 10^{-6}} \cdot 4 \times 10^{-6} = 2 \, \text{m} \]

### Step 6: Final calculation
The total distance \(x\) that the man can see is approximately:
\[ x = 2000 \, \text{m} = 2 \, \text{km} \]

### Conclusion
The actual length of the road that the man is able to see is **2000 meters** or **2 kilometers**.

---