Wavelength of blue in vacuum is `5000 Å`. What will be the thickness of an air column of same thickness which have 20 more wavelengths of blue light ? Given the refractive index of air column is 1.003.

Answer

To solve the problem step by step, we will use the information provided about the wavelength of blue light in vacuum and the refractive index of air.

### Step 1: Understand the relationship between wavelength, thickness, and number of wavelengths
The thickness of a medium can be expressed in terms of the wavelength of light in that medium and the number of wavelengths present.

For a medium (like vacuum), the thickness \( L_v \) can be given by:
\[ L_v = \lambda_v \times n_1 \]
where \( \lambda_v \) is the wavelength in vacuum and \( n_1 \) is the number of wavelengths in that thickness.

For the air column, the thickness \( L_a \) can be expressed as:
\[ L_a = \lambda_a \times n_2 \]
where \( \lambda_a \) is the wavelength in air and \( n_2 \) is the number of wavelengths in that thickness.

### Step 2: Relate the wavelengths in vacuum and air
The wavelength of light in air can be calculated using the formula:
\[ \lambda_a = \frac{\lambda_v}{n} \]
where \( n \) is the refractive index of air. Given that the refractive index of air is \( 1.003 \), we can write:
\[ \lambda_a = \frac{5000 \, \text{Å}}{1.003} \]

### Step 3: Calculate the wavelength in air
Calculating \( \lambda_a \):
\[ \lambda_a = \frac{5000 \, \text{Å}}{1.003} \approx 4985.07 \, \text{Å} \]

### Step 4: Set up the equation for thickness
According to the problem, we need to find the thickness of the air column that has 20 more wavelengths than the same thickness in vacuum. Therefore, we can write:
\[ L_v = \lambda_v \times n_1 \]
\[ L_a = \lambda_a \times (n_1 + 20) \]

### Step 5: Equate the two expressions for thickness
Since \( L_v = L_a \), we can set the two equations equal to each other:
\[ \lambda_v \times n_1 = \lambda_a \times (n_1 + 20) \]

### Step 6: Substitute the values
Substituting the known values:
\[ 5000 \times n_1 = 4985.07 \times (n_1 + 20) \]

### Step 7: Solve for \( n_1 \)
Expanding the right side:
\[ 5000 n_1 = 4985.07 n_1 + 99701.4 \]
Rearranging gives:
\[ 5000 n_1 - 4985.07 n_1 = 99701.4 \]
\[ 14.93 n_1 = 99701.4 \]
\[ n_1 \approx \frac{99701.4}{14.93} \approx 6675.23 \]

### Step 8: Calculate the thickness in air
Now we can find the thickness of the air column:
\[ L_a = \lambda_a \times (n_1 + 20) \]
Substituting \( n_1 \):
\[ L_a = 4985.07 \times (6675.23 + 20) \]
Calculating this gives:
\[ L_a \approx 4985.07 \times 6695.23 \approx 33412.5 \, \text{Å} \]

### Step 9: Convert to mm
To convert this to mm, we divide by \( 10^7 \):
\[ L_a \approx \frac{33412.5}{10^7} \approx 0.00334125 \, \text{mm} \approx 3.34 \, \text{mm} \]

### Final Answer
The thickness of the air column is approximately **3.34 mm**.

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