A rectangular tank filled with some liquid is accelerated along a horizontal surface at `(40)/(3)ms^(-2)`. Inside the liquid, a laser pointer is fixed a the centre of the tank which shoots a thin laser beam in the vertically upward direction. If after refraction from the liquid surface, the laser beam moves along the surface of the liquid, then what is the refractive index of the liquid ?

Answer

To solve the problem, we need to determine the refractive index of the liquid in the tank based on the given conditions. Here are the steps to arrive at the solution:

### Step 1: Understand the Forces Acting on the Liquid
When the tank is accelerated horizontally, a pseudo force acts on the liquid in the opposite direction of the acceleration. The gravitational force acts downwards. The effective force acting on the liquid can be represented as a combination of these two forces.

### Step 2: Determine the Angle of the Liquid Surface
The angle θ of the liquid surface can be found using the relationship between the gravitational force (g) and the pseudo force (a). The pseudo force is equal to the acceleration of the tank, which is given as \( \frac{40}{3} \, \text{m/s}^2 \).

Using the tangent of the angle:
\[ \tan \theta = \frac{\text{pseudo force}}{\text{gravitational force}} = \frac{a}{g} \]
Substituting the values:
\[ \tan \theta = \frac{\frac{40}{3}}{10} = \frac{4}{3} \]

### Step 3: Calculate the Angle θ
From the tangent value, we can find θ:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Apply Snell's Law
When the laser beam passes from the liquid to the air, Snell's law applies:
\[ n_1 \sin i = n_2 \sin r \]
Where:
- \( n_1 \) is the refractive index of the liquid (to be determined),
- \( n_2 = 1 \) (the refractive index of air),
- \( i \) is the angle of incidence (which is equal to θ),
- \( r \) is the angle of refraction (which is 90° since the beam moves along the surface of the liquid).

Since \( \sin 90° = 1 \), we can rewrite Snell's law as:
\[ n_1 \sin \theta = 1 \]

### Step 5: Calculate the Sine of θ
From the triangle formed by the forces, we can find \( \sin \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{\sqrt{4^2 + 3^2}} = \frac{4}{5} \]

### Step 6: Substitute into Snell's Law
Substituting \( \sin \theta \) into Snell's law:
\[ n_1 \cdot \frac{4}{5} = 1 \]
Thus,
\[ n_1 = \frac{5}{4} \]

### Step 7: Final Answer
The refractive index of the liquid is:
\[ n_1 = 1.25 \]

### Summary
The refractive index of the liquid in the tank is 1.25.

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