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 refrctive 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’s the step-by-step solution:

### Step 1: Understand the Situation
The tank is accelerating horizontally with an acceleration \( a = \frac{40}{3} \, \text{m/s}^2 \). A laser pointer shoots a beam vertically upward, and after refraction at the liquid surface, the beam travels along the surface of the liquid.

### Step 2: Analyze Forces in the Accelerating Frame
In the accelerating frame of the liquid, we consider the effects of pseudo force. The forces acting on a small mass \( m \) of the liquid are:
- Gravitational force \( mg \) acting downward.
- Pseudo force \( ma \) acting horizontally in the opposite direction of acceleration.

### Step 3: Establish the Angle of the Liquid Surface
At equilibrium, the forces must balance. The angle \( \theta \) that the surface of the liquid makes with the horizontal can be determined using:
\[ m a \cos \theta = mg \sin \theta \]
This simplifies to:
\[ \tan \theta = \frac{a}{g} \]
Substituting the given values:
\[ \tan \theta = \frac{\frac{40}{3}}{10} = \frac{4}{3} \]
Thus, \( \theta = \tan^{-1}(\frac{4}{3}) \).

### Step 4: Calculate the Angle \( \theta \)
Using trigonometric values, we find:
\[ \theta \approx 53^\circ \]

### Step 5: Refraction at the Liquid Surface
When the laser beam refracts at the liquid surface, it travels along the surface after refraction. According to Snell's Law:
\[ \mu \sin i = n \sin r \]
Where:
- \( \mu \) is the refractive index of the liquid,
- \( i \) is the angle of incidence (which is \( \theta \)),
- \( n \) is the refractive index of air (approximately 1),
- \( r \) is the angle of refraction (which is \( 90^\circ \)).

Since \( \sin 90^\circ = 1 \), we can rewrite Snell's Law as:
\[ \mu \sin \theta = 1 \cdot 1 \]
Thus:
\[ \mu = \frac{1}{\sin \theta} \]

### Step 6: Calculate \( \sin \theta \)
From trigonometric ratios, we know:
\[ \sin 53^\circ = \frac{4}{5} \]
Therefore:
\[ \mu = \frac{1}{\frac{4}{5}} = \frac{5}{4} = 1.25 \]

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