A man of height `2.0 m` is standing on a level road because of temprature variation the refractive index of air is varying as `mu=sqrt(1+ay),` where y is height from road. If `a=2.0xx10^-6 m^-1.` Then, find distant point that he can see on the road.

Answer

To solve the problem step by step, we will analyze the situation using the given information about the refractive index and the geometry involved.

### Step 1: Understanding the Refractive Index Variation
The refractive index of air is given by:
\[ \mu = \sqrt{1 + ay} \]
where \( a = 2.0 \times 10^{-6} \, \text{m}^{-1} \) and \( y \) is the height from the road.

### Step 2: Setting Up the Geometry
We have a man of height \( h = 2.0 \, \text{m} \) standing on a level road. We need to find the maximum distance \( x \) that he can see on the road. The light rays coming from a distant point on the road will bend due to the varying refractive index.

### Step 3: Applying Snell's Law
According to Snell's law:
\[ \mu_0 \sin \theta = \mu \sin 90^\circ \]
Here, \( \mu_0 = 1 \) (for air) and \( \sin 90^\circ = 1 \). Thus,
\[ \sin \theta = \frac{1}{\mu} \]
Substituting the expression for \( \mu \):
\[ \sin \theta = \frac{1}{\sqrt{1 + ay}} \]

### Step 4: Relating \( \tan \theta \) and Geometry
Using basic trigonometric functions, we have:
\[ \tan \theta = \frac{dx}{dy} \]
From the previous step:
\[ \tan \theta = \frac{1}{\sqrt{1 + ay}} \]
Thus, we can equate:
\[ \frac{dx}{dy} = \frac{1}{\sqrt{1 + ay}} \]

### Step 5: Rearranging and Integrating
Rearranging gives us:
\[ dx = \frac{dy}{\sqrt{1 + ay}} \]
Now we will integrate both sides. The limits for \( y \) will be from \( 0 \) to \( 2 \, \text{m} \) (the height of the man), and we want to find \( x \) when \( y = 2 \):
\[ \int_0^x dx = \int_0^2 \frac{dy}{\sqrt{1 + ay}} \]

### Step 6: Solving the Integral
The integral on the right can be solved using a substitution. Let \( u = 1 + ay \), then \( du = a \, dy \) or \( dy = \frac{du}{a} \). The limits change accordingly:
- When \( y = 0 \), \( u = 1 \)
- When \( y = 2 \), \( u = 1 + 2a \)

Thus, the integral becomes:
\[ x = \int_1^{1 + 2a} \frac{1}{\sqrt{u}} \frac{du}{a} = \frac{1}{a} \left[ 2\sqrt{u} \right]_1^{1 + 2a} \]
Calculating this gives:
\[ x = \frac{2}{a} \left( \sqrt{1 + 2a} - 1 \right) \]

### Step 7: Substituting the Value of \( a \)
Now, substituting \( a = 2.0 \times 10^{-6} \):
\[ x = \frac{2}{2.0 \times 10^{-6}} \left( \sqrt{1 + 4.0 \times 10^{-6}} - 1 \right) \]
Calculating \( \sqrt{1 + 4.0 \times 10^{-6}} \approx 1 + 2.0 \times 10^{-6} \) (using the binomial approximation for small values):
\[ x \approx \frac{2}{2.0 \times 10^{-6}} \left( 2.0 \times 10^{-6} \right) = 2000 \, \text{m} \]

### Final Result
The maximum distance that the man can see on the road is:
\[ \boxed{2000 \, \text{m}} \]