In Young's double slit experiment, angular width of fringes is `0.20^(@)` for sodium light of wavelength 5890 Å. If the complete system is dipped in water, then the angular width of fringes will become

Options

A. `0.11^(@)`
B. `0.15^(@)`
C. `0.22^(@)`
D. `0.30^(@)`

Explanation

To solve the problem of finding the new angular width of fringes when the Young's double slit experiment is conducted in water, we can follow these steps:

### Step 1: Understand the relationship between angular width and wavelength
In Young's double slit experiment, the angular width (θ) of the fringes is given by the formula:
\[ \theta = \frac{\lambda}{d} \]
where:
- \( \theta \) is the angular width,
- \( \lambda \) is the wavelength of the light,
- \( d \) is the distance between the slits.

### Step 2: Determine the initial conditions
We are given:
- The initial angular width in air (θ) = 0.20 degrees,
- The wavelength of sodium light (λ) = 5890 Å (which is \( 5890 \times 10^{-10} \) m).

### Step 3: Understand the effect of the medium
When the entire system is dipped in water, the wavelength of light changes due to the refractive index of water (μ). The new wavelength in water (λ') can be calculated using:
\[ \lambda' = \frac{\lambda}{\mu} \]
where \( \mu \) for water is approximately \( \frac{4}{3} \).

### Step 4: Calculate the new wavelength in water
Substituting the values:
\[ \lambda' = \frac{5890 \, \text{Å}}{\frac{4}{3}} = 5890 \, \text{Å} \times \frac{3}{4} = 4417.5 \, \text{Å} \]

### Step 5: Calculate the new angular width in water
Now, we can find the new angular width (θ') in water using the relationship:
\[ \theta' = \frac{\lambda'}{d} \]
Since \( d \) remains constant, we can express the new angular width in terms of the old angular width:
\[ \theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta \cdot d}{\mu d} = \frac{\theta}{\mu} \]
Thus:
\[ \theta' = \frac{0.20 \, \text{degrees}}{\frac{4}{3}} = 0.20 \times \frac{3}{4} = 0.15 \, \text{degrees} \]

### Conclusion
The new angular width of the fringes when the entire system is dipped in water is:
\[ \theta' = 0.15 \, \text{degrees} \]