In a Young's double slit experiment, an angular width of the fringe is `0.35^(@)` on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe. When whole system is immersed in a medium of refractive index 7/5 is `(1)/(alpha)`. The value of `alpha` is ________

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Angular width in air, \( \theta_0 = 0.35^\circ \)
- Distance from the slits to the screen, \( D = 2 \, \text{m} \)
- Wavelength of light, \( \lambda = 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m} \)
- Refractive index of the medium, \( \mu = \frac{7}{5} \)

### Step 2: Convert angular width from degrees to radians
To perform calculations, we need to convert the angular width from degrees to radians:
\[ \theta_0 = 0.35^\circ \times \frac{\pi}{180^\circ} = \frac{0.35 \times \pi}{180} \, \text{radians} \]

### Step 3: Calculate the new angular width in the medium
In a medium with refractive index \( \mu \), the angular width \( \theta \) is given by:
\[ \theta = \frac{\theta_0}{\mu} \]
Substituting the values:
\[ \theta = \frac{0.35^\circ}{\frac{7}{5}} = 0.35^\circ \times \frac{5}{7} \]

### Step 4: Calculate the numerical value of the new angular width
Calculating the new angular width:
\[ \theta = 0.35 \times \frac{5}{7} = \frac{1.75}{7} = 0.25^\circ \]

### Step 5: Relate the new angular width to \( \alpha \)
According to the problem, the new angular width is given as:
\[ \theta = \frac{1}{\alpha} \]
From our calculation, we have:
\[ 0.25^\circ = \frac{1}{\alpha} \]

### Step 6: Solve for \( \alpha \)
Taking the reciprocal of both sides gives:
\[ \alpha = \frac{1}{0.25} = 4 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 4 \]

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