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

Options

A. (a) `0.11^@`
B. (b) `0.15^@`
C. (c) `0.22^@`
D. (d) `0.30^@`

Explanation

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

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

### Step 2: Determine the effect of the medium on the wavelength.
When the system is dipped in water, the wavelength of light changes. The new wavelength (\( \lambda_{water} \)) in the medium can be calculated using the refractive index (\( \mu \)) of water:
\[ \lambda_{water} = \frac{\lambda_{air}}{\mu} \]
For water, the refractive index \( \mu \) is approximately \( \frac{4}{3} \).

### Step 3: Substitute the values.
Given:
- \( \lambda_{air} = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \)
- \( \mu_{water} = \frac{4}{3} \)

Now, calculate \( \lambda_{water} \):
\[ \lambda_{water} = \frac{5890 \times 10^{-10}}{\frac{4}{3}} = \frac{5890 \times 10^{-10} \times 3}{4} = \frac{17670 \times 10^{-10}}{4} = 4417.5 \times 10^{-10} \, \text{m} \]

### Step 4: Calculate the new angular width in water.
Using the new wavelength in water, the angular width in water (\( \theta_{water} \)) can be calculated as:
\[ \theta_{water} = \frac{\lambda_{water}}{d} \]
Since \( d \) remains constant, we can express \( \theta_{water} \) in terms of \( \theta_{air} \):
\[ \theta_{water} = \frac{\lambda_{water}}{d} = \frac{\lambda_{air}/\mu}{d} = \frac{\lambda_{air}}{d \cdot \mu} = \frac{\theta_{air}}{\mu} \]

### Step 5: Substitute the known values.
Given that \( \theta_{air} = 0.20^\circ \):
\[ \theta_{water} = \frac{0.20^\circ}{\frac{4}{3}} = 0.20^\circ \times \frac{3}{4} = 0.15^\circ \]

### Final Answer:
The angular width of the fringes when the system is dipped in water is \( 0.15^\circ \).

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