Calculate the wavelength of light used in an interference experiment from the following data: Fringe width = 0.03 cm. Distance between slits and eyepiece through which the interference pattern is observed is 1 m. Distance between the images of the virtural when a convex lens of focal length 16 cm is used at a distance of 80 cm from the eyepiece is 0.8 cm.

Options

A. `6000 Å`
B. `0.00006 Å`
C. 6000 cm
D. 0.00006 m

Explanation

To calculate the wavelength of light used in the interference experiment, we will follow these steps:

### Step 1: Convert Given Values
- Fringe width \( \beta = 0.03 \, \text{cm} \)
- Distance between slits and eyepiece \( D = 1 \, \text{m} = 100 \, \text{cm} \)
- Distance between images \( d = 0.8 \, \text{cm} \)
- Focal length of the lens \( f = 16 \, \text{cm} \)
- Distance of the lens from the eyepiece \( v = 80 \, \text{cm} \)

### Step 2: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Substituting the known values:
\[ \frac{1}{80} + \frac{1}{u} = \frac{1}{16} \]

### Step 3: Solve for \( u \)
Rearranging the equation to find \( u \):
\[ \frac{1}{u} = \frac{1}{16} - \frac{1}{80} \]
Finding a common denominator (80):
\[ \frac{1}{u} = \frac{5 - 1}{80} = \frac{4}{80} = \frac{1}{20} \]
Thus, \( u = 20 \, \text{cm} \).

### Step 4: Calculate Magnification
The magnification \( M \) is given by:
\[ M = \frac{v}{u} = \frac{80}{20} = 4 \]

### Step 5: Relate Magnification to Fringe Width
The magnification can also be expressed in terms of the distances:
\[ M = \frac{d}{D} \]
Substituting the values:
\[ 4 = \frac{0.8}{D} \]
Solving for \( D \):
\[ D = \frac{0.8}{4} = 0.2 \, \text{cm} \]

### Step 6: Use the Fringe Width Formula
The fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the known values:
\[ 0.03 = \frac{\lambda \cdot 100}{0.2} \]

### Step 7: Solve for Wavelength \( \lambda \)
Rearranging the equation:
\[ \lambda = \frac{0.03 \cdot 0.2}{100} \]
Calculating \( \lambda \):
\[ \lambda = \frac{0.006}{100} = 0.00006 \, \text{cm} = 6000 \, \text{Å} \]

### Final Answer
The wavelength of light used in the interference experiment is \( \lambda = 6000 \, \text{Å} \).

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