To solve the problem step by step, we will follow the outlined procedure in the video transcript and add necessary calculations.
### Step 1: Understand the Setup
We have a convergent lens with a focal length \( f = 10 \, \text{cm} \) that is cut into two halves and separated by a distance \( d = 0.5 \, \text{mm} = 0.05 \, \text{cm} \). A point source of monochromatic light with a wavelength \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \) is placed at a distance \( a = 15 \, \text{cm} \) from the lens. We need to find the fringe width on a screen located \( D = 60 \, \text{cm} \) behind the lens.
### Step 2: Find the Image Distance \( v \)
Using the lens formula:
\[
\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
\]
where \( u = -15 \, \text{cm} \) (negative because the object is on the same side as the incoming light).
Substituting the values:
\[
\frac{1}{v} - \frac{1}{-15} = \frac{1}{10}
\]
This simplifies to:
\[
\frac{1}{v} + \frac{1}{15} = \frac{1}{10}
\]
Now, finding a common denominator (which is 30):
\[
\frac{2}{30} + \frac{2}{30} = \frac{3}{30}
\]
Thus:
\[
\frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}
\]
So:
\[
v = 30 \, \text{cm}
\]
### Step 3: Calculate the Effective Distance Between the Lens Halves
The distance between the two halves of the lens is \( d = 0.05 \, \text{cm} \). Since the lens is cut into two halves, the distance between the two effective sources (the two halves of the lens) is \( d = 0.05 \, \text{cm} \).
### Step 4: Calculate the Total Distance to the Screen
The distance from the lens to the screen is given as \( D = 60 \, \text{cm} \). The distance from the image formed to the screen is:
\[
D' = D - v = 60 \, \text{cm} - 30 \, \text{cm} = 30 \, \text{cm}
\]
### Step 5: Calculate the Fringe Width
The formula for fringe width \( w \) is given by:
\[
w = \frac{\lambda D'}{d}
\]
Substituting the values:
\[
w = \frac{5 \times 10^{-7} \, \text{m} \times 30 \times 10^{-2} \, \text{m}}{0.05 \times 10^{-2} \, \text{m}}
\]
Calculating this:
\[
w = \frac{5 \times 10^{-7} \times 30 \times 10^{-2}}{0.05 \times 10^{-2}} = \frac{5 \times 30}{0.05} \times 10^{-7} = \frac{150}{0.05} \times 10^{-7} = 3000 \times 10^{-7} = 3 \times 10^{-4} \, \text{m} = 0.3 \, \text{mm}
\]
### Final Answer
The fringe width on the screen is \( 0.3 \, \text{mm} \).
