Light from a sodium lamp `(lambda_0=589 nm)` passes through a tank of glycerin (refractive index `=1.47`) `20m` long in a time `t_1.` If it takes a time `t_2` to transverse the same tank when filled with carbon disulfide (index`=1.63`), determine the difference `t_2-t_1.`

Explanation

To solve the problem of finding the difference in time taken by light to traverse a 20 m long tank filled with glycerin and carbon disulfide, we can follow these steps:

### Step 1: Understand the relationship between speed, distance, and time
The time taken by light to travel a certain distance can be expressed using the formula:
\[ t = \frac{s}{v} \]
where \( t \) is the time, \( s \) is the distance, and \( v \) is the speed of light in the medium.

### Step 2: Calculate the speed of light in glycerin
The speed of light in a medium is given by:
\[ v = \frac{c}{\mu} \]
where \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \) m/s) and \( \mu \) is the refractive index of the medium.

For glycerin, with a refractive index \( \mu_1 = 1.47 \):
\[ v_1 = \frac{c}{\mu_1} = \frac{3 \times 10^8 \text{ m/s}}{1.47} \]

### Step 3: Calculate the time \( t_1 \) for glycerin
Using the distance \( s = 20 \) m:
\[ t_1 = \frac{s}{v_1} = \frac{20 \text{ m}}{v_1} \]

### Step 4: Calculate the speed of light in carbon disulfide
For carbon disulfide, with a refractive index \( \mu_2 = 1.63 \):
\[ v_2 = \frac{c}{\mu_2} = \frac{3 \times 10^8 \text{ m/s}}{1.63} \]

### Step 5: Calculate the time \( t_2 \) for carbon disulfide
Using the same distance \( s = 20 \) m:
\[ t_2 = \frac{s}{v_2} = \frac{20 \text{ m}}{v_2} \]

### Step 6: Find the difference \( t_2 - t_1 \)
Now we can find the difference in time:
\[ t_2 - t_1 = \frac{20 \text{ m}}{v_2} - \frac{20 \text{ m}}{v_1} \]
Factoring out \( 20 \text{ m} \):
\[ t_2 - t_1 = 20 \text{ m} \left( \frac{1}{v_2} - \frac{1}{v_1} \right) \]

### Step 7: Substitute \( v_1 \) and \( v_2 \) into the equation
Substituting the values of \( v_1 \) and \( v_2 \):
\[ t_2 - t_1 = 20 \text{ m} \left( \frac{1.47}{3 \times 10^8} - \frac{1.63}{3 \times 10^8} \right) \]

### Step 8: Simplify the expression
This simplifies to:
\[ t_2 - t_1 = 20 \text{ m} \cdot \frac{(1.47 - 1.63)}{3 \times 10^8} \]
\[ t_2 - t_1 = 20 \text{ m} \cdot \frac{-0.16}{3 \times 10^8} \]

### Step 9: Calculate the final result
Calculating the difference:
\[ t_2 - t_1 = 20 \cdot \frac{-0.16}{3 \times 10^8} = \frac{-3.2}{3 \times 10^8} \]
\[ t_2 - t_1 = -1.0667 \times 10^{-8} \text{ s} \]

### Final Answer
The difference \( t_2 - t_1 \) is approximately:
\[ t_2 - t_1 \approx -1.067 \times 10^{-8} \text{ seconds} \]