A monchromatic beam of electrons accelerated by a potential difference V falls normally on the plate containing two narrow slits seperated by a distance d. The interference pattern is observed on a screen parallel to the plane of the slit and at a distance of D from slits. Fringe width is found to be `omega_(1)`. When electron beam is accelerated by the potential differene of 4V, the finge width becomes `omega_(2)`. Find the ratio `(omega_(1))/(omega_(2))`. (Given `dltltD`)

Answer

To solve the problem step by step, we will analyze the relationship between the fringe width and the potential difference applied to the electrons in the double-slit experiment.

### Step 1: Understand the relationship between fringe width and wavelength
The fringe width (ω) in a double-slit experiment is given by the formula:
\[ \omega = \frac{\lambda D}{d} \]
where:
- \( \lambda \) is the wavelength of the electrons,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 2: Express the fringe widths for two different potential differences
When the electrons are accelerated by a potential difference \( V \), the fringe width is \( \omega_1 \):
\[ \omega_1 = \frac{\lambda_1 D}{d} \]
When the potential difference is increased to \( 4V \), the fringe width becomes \( \omega_2 \):
\[ \omega_2 = \frac{\lambda_2 D}{d} \]

### Step 3: Relate the wavelength to the potential difference
The wavelength \( \lambda \) of the electrons can be expressed in terms of the potential difference \( V \) using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be related to the kinetic energy, which is given by:
\[ KE = eV = \frac{p^2}{2m} \]
From this, we can derive:
\[ p = \sqrt{2m eV} \]
Substituting this into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]
Thus, we can see that:
\[ \lambda \propto \frac{1}{\sqrt{V}} \]

### Step 4: Express the wavelengths for the two potential differences
Using the relationship derived above:
\[ \lambda_1 \propto \frac{1}{\sqrt{V}} \quad \text{and} \quad \lambda_2 \propto \frac{1}{\sqrt{4V}} = \frac{1}{2\sqrt{V}} \]

### Step 5: Find the ratio of the wavelengths
Now we can find the ratio of the wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\sqrt{V}}}{\frac{1}{2\sqrt{V}}} = \frac{1}{\sqrt{V}} \cdot \frac{2\sqrt{V}}{1} = 2 \]

### Step 6: Find the ratio of fringe widths
Since the fringe widths are directly proportional to the wavelengths:
\[ \frac{\omega_1}{\omega_2} = \frac{\lambda_1}{\lambda_2} = 2 \]

### Final Answer
Thus, the ratio of the fringe widths is:
\[ \frac{\omega_1}{\omega_2} = 2 \]