A double slit arrangement produces fringes for light `lambda=5890Å` which are `0.2^@` apart. If the whole arrangement is fully dipped in a liquid of `mu=4/3`, the angular fringes separation is

Options

A. (a) `0.15^@`
B. (b) `0.30^@`
C. (c) `0.10^@`
D. (d) `0.25^@`

Explanation

To solve the problem, we need to find the angular fringe separation after the double slit arrangement is dipped in a liquid with a refractive index (μ) of 4/3. Let's break down the solution step by step.

### Step 1: Understand the given data
- Wavelength of light, \( \lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, \text{m} \)
- Initial angular fringe separation, \( \theta_1 = 0.2^\circ \)
- Refractive index of the liquid, \( \mu = \frac{4}{3} \)

### Step 2: Calculate the new wavelength in the liquid
When the setup is dipped in a liquid, the wavelength of light changes according to the formula:
\[ \lambda' = \frac{\lambda}{\mu} \]
Substituting the values:
\[ \lambda' = \frac{5890 \times 10^{-10}}{\frac{4}{3}} = 5890 \times 10^{-10} \times \frac{3}{4} = \frac{17670 \times 10^{-10}}{4} = 4417.5 \times 10^{-10} \, \text{m} \]

### Step 3: Relate the angular fringe separations
The angular fringe separation is proportional to the wavelength. Thus, we can write:
\[ \frac{\theta_1}{\theta_2} = \frac{\lambda}{\lambda'} \]
Where:
- \( \theta_1 = 0.2^\circ \) (initial fringe separation)
- \( \theta_2 \) is the new fringe separation we want to find.

### Step 4: Substitute the values into the equation
Now substituting the values we have:
\[ \frac{0.2^\circ}{\theta_2} = \frac{\lambda}{\lambda'} = \frac{\lambda}{\frac{\lambda}{\mu}} = \mu \]
So,
\[ \theta_2 = \frac{0.2^\circ}{\mu} = \frac{0.2^\circ}{\frac{4}{3}} = 0.2^\circ \times \frac{3}{4} = 0.15^\circ \]

### Step 5: Conclusion
Thus, the angular fringe separation after dipping the arrangement in the liquid is:
\[ \theta_2 = 0.15^\circ \]

### Final Answer
The angular fringe separation is \( 0.15^\circ \).

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