While determining the refractive index of a liquid experimentally, the microscope was focussed at the bottom of a beaker, when its reading was 3.965 cm. on pouring liquid upto a height 2.537 cm inside the beaker, the reading of the refocussed microscope was 3.348 cm. Find the refractive index of the liquid.

Answer

To find the refractive index of the liquid, we can follow these steps:

### Step 1: Identify the initial and final readings
- The initial reading of the microscope when focused at the bottom of the beaker is \( h_1 = 3.965 \, \text{cm} \).
- After pouring the liquid up to a height of \( h = 2.537 \, \text{cm} \), the new reading of the microscope is \( h_2 = 3.348 \, \text{cm} \).

### Step 2: Calculate the change in height
- The change in height (\( \Delta h \)) due to the liquid is given by:
\[ \Delta h = h_1 - h_2 = 3.965 \, \text{cm} - 3.348 \, \text{cm} \]
- Performing the calculation:
\[ \Delta h = 0.617 \, \text{cm} \]

### Step 3: Relate the change in height to the refractive index
- The relationship between the height change and the refractive index (\( \mu \)) is given by:
\[ \Delta h = T \left( 1 - \frac{1}{\mu} \right) \]
where \( T \) is the height of the liquid column, which is \( T = 2.537 \, \text{cm} \).

### Step 4: Substitute the known values into the equation
- Rearranging the equation gives:
\[ \frac{1}{\mu} = 1 - \frac{\Delta h}{T} \]
- Substituting \( \Delta h = 0.617 \, \text{cm} \) and \( T = 2.537 \, \text{cm} \):
\[ \frac{1}{\mu} = 1 - \frac{0.617}{2.537} \]

### Step 5: Calculate \( \frac{0.617}{2.537} \)
- Performing the division:
\[ \frac{0.617}{2.537} \approx 0.243 \]
- Therefore:
\[ \frac{1}{\mu} = 1 - 0.243 = 0.757 \]

### Step 6: Calculate the refractive index \( \mu \)
- Taking the reciprocal gives:
\[ \mu = \frac{1}{0.757} \approx 1.321 \]

### Final Answer
- The refractive index of the liquid is approximately \( \mu \approx 1.321 \).

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