Question

Because of pressure and temperature difference the refractive index of air varies with height $h$ as $\mu = \mu_0\sqrt{1+(10^{-6})h}$. There is a point object placed on surface, the maximum horizontal distance of the observer at height 4m is able to see it is $d$ km

• A. The angle with horizontal the point object will be seen by observer is $\theta = \tan^{-1}(10^{-2})$ rad.
• B. $d = 4$ km
• C. Horizontal distance of point on trajectory is 2 km for height of 1 m
• D. Equation is trajectory is hyperbolic in nature

Answer 1

Answer: (B), (C)

B, C

Answer 2

The problem describes the variation of the refractive index of air with height $h$ as $\mu = \mu_0\sqrt{1+(10^{-6})h}$. We need to analyze the path of light rays and determine the maximum horizontal distance an observer at a certain height can see an object on the surface.

1. Principle of Light Ray Propagation in a Stratified Medium:

For a light ray propagating in a medium where the refractive index $\mu$ varies only with height $h$, Snell's law in differential form implies that the quantity $\mu \cos \theta$ remains constant, where $\theta$ is the angle the ray makes with the horizontal. Let the light ray start from the object placed on the surface ($h=0$) with an initial angle $\theta_0$ with the horizontal. So, $\mu(h) \cos \theta = \mu_0 \cos \theta_0$.

2. Condition for Maximum Horizontal Distance:

To achieve the maximum horizontal distance for an observer at height $H$ to see an object on the surface, the light ray must be launched from the object at the smallest possible angle with the horizontal, effectively tangent to the surface. This means $\theta_0 \to 0$, so $\cos \theta_0 \to 1$. Under this condition, the constant becomes $\mu_0$. Thus, the relation for the ray that defines the maximum horizontal distance is: $\mu_0\sqrt{1+(10^{-6})h} \cos \theta = \mu_0$ $\cos \theta = \frac{1}{\sqrt{1+(10^{-6})h}}$

3. Trajectory of the Light Ray:

The horizontal distance $x$ traversed by the ray for a vertical change $dh$ is given by $dx = \frac{dh}{\tan \theta}$. First, let's find $\tan \theta$: $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{1+(10^{-6})h} = \frac{1+(10^{-6})h - 1}{1+(10^{-6})h} = \frac{(10^{-6})h}{1+(10^{-6})h}$ $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{\frac{(10^{-6})h}{1+(10^{-6})h}}}{\frac{1}{\sqrt{1+(10^{-6})h}}} = \sqrt{(10^{-6})h}$

Now, substitute $\tan \theta$ into the $dx$ equation: $dx = \frac{dh}{\sqrt{(10^{-6})h}} = \frac{1}{\sqrt{10^{-6}}} \frac{dh}{\sqrt{h}} = 10^3 h^{-1/2} dh$

To find the horizontal distance $x$ for a given height $h$, we integrate from $0$ to $h$: $x = \int_0^h 10^3 h'^{-1/2} dh' = 10^3 \left[ \frac{h'^{1/2}}{1/2} \right]_0^h = 10^3 [2\sqrt{h'}]_0^h = 2 \times 10^3 \sqrt{h}$ So, the equation of the trajectory is $x = 2 \times 10^3 \sqrt{h}$.

4. Evaluating the Options:

A. The angle with horizontal the point object will be seen by observer is $\theta = \tan^{-1}(10^{-2})$ rad. The observer is at height $H=4$m. The angle $\theta$ is the angle the ray makes with the horizontal at the observer's position. Using $\tan \theta = \sqrt{(10^{-6})h}$ and $h=4$m: $\tan \theta = \sqrt{(10^{-6}) \times 4} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3}$ So, $\theta = \tan^{-1}(2 \times 10^{-3})$ rad. Since $2 \times 10^{-3} = 0.002$ and $10^{-2} = 0.01$, option A is incorrect.

B. $d = 4$ km $d$ is the maximum horizontal distance when the observer is at height $H=4$m. Using the trajectory equation $x = 2 \times 10^3 \sqrt{h}$, set $h=4$m: $d = 2 \times 10^3 \sqrt{4} = 2 \times 10^3 \times 2 = 4 \times 10^3$ meters $= 4$ km. So, option B is correct.

C. Horizontal distance of point on trajectory is 2 km for height of 1 m Using the trajectory equation $x = 2 \times 10^3 \sqrt{h}$, set $h=1$m: $x = 2 \times 10^3 \sqrt{1} = 2 \times 10^3$ meters $= 2$ km. So, option C is correct.

D. Equation is trajectory is hyperbolic in nature The equation of the trajectory is $x = 2 \times 10^3 \sqrt{h}$. Squaring both sides, we get $x^2 = (2 \times 10^3)^2 h = 4 \times 10^6 h$. Rearranging, $h = \frac{1}{4 \times 10^6} x^2$. This is of the form $y = ax^2$, which represents a parabola, not a hyperbola. So, option D is incorrect.

The final answer is $\boxed{B, C}$

Explanation of the solution:

1. Snell's Law for Varying Refractive Index: For light traveling in a medium where refractive index $\mu$ varies only with height $h$, the quantity $\mu \cos \theta$ (where $\theta$ is the angle with the horizontal) remains constant.

2. Maximum Horizontal Distance: To achieve the maximum horizontal distance for an observer at height $H$ to see an object on the ground, the light ray must be launched from the object horizontally ($\theta_0 \approx 0^\circ$). This makes the constant $\mu_0 \cos 0^\circ = \mu_0$.

3. Ray Path Equation: From $\mu \cos \theta = \mu_0$, substitute $\mu = \mu_0\sqrt{1+(10^{-6})h}$ to get $\cos \theta = \frac{1}{\sqrt{1+(10^{-6})h}}$.

4. Differential Horizontal Distance: The change in horizontal distance $dx$ for a change in height $dh$ is $dx = \frac{dh}{\tan \theta}$. Using $\tan \theta = \sqrt{\frac{1}{\cos^2 \theta} - 1} = \sqrt{1+(10^{-6})h - 1} = \sqrt{(10^{-6})h}$.

5. Integration for Trajectory: Integrate $dx = \frac{dh}{\sqrt{(10^{-6})h}} = 10^3 h^{-1/2} dh$ from $0$ to $h$ to get the trajectory equation $x = 2 \times 10^3 \sqrt{h}$.

6. Verify Options:

   • A (Angle): At $h=4$m, $\tan \theta = \sqrt{(10^{-6}) \times 4} = 2 \times 10^{-3}$. So $\theta = \tan^{-1}(2 \times 10^{-3})$, not $\tan^{-1}(10^{-2})$. Incorrect.
   • B (Distance $d$): For $H=4$m, $d = 2 \times 10^3 \sqrt{4} = 4 \times 10^3$ m $= 4$ km. Correct.
   • C (Distance at 1m): For $h=1$m, $x = 2 \times 10^3 \sqrt{1} = 2 \times 10^3$ m $= 2$ km. Correct.
   • D (Trajectory Type): Squaring $x = 2 \times 10^3 \sqrt{h}$ gives $x^2 = 4 \times 10^6 h$, or $h = \frac{1}{4 \times 10^6} x^2$. This is a parabolic equation ($y=ax^2$). Incorrect.

Answer: The correct options are B and C.