A double slit arrangement produces interference fringes for sodium light `(lambda=589nm)` that have an angular separation of `3.50xx10^-3` radian. For what wavelength would the angular separation be 10% greater?

Options

A. (a) `527nm`
B. (b) `648nm`
C. (c) `722nm`
D. (d) `449nm`

Explanation

To solve the problem, we need to determine the new wavelength when the angular separation is 10% greater than the original angular separation produced by sodium light. Let's break this down step by step.

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

### Step 2: Calculate the original angular separation
From the problem, we know:
- Original wavelength, \(\lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m}\)
- Original angular separation, \(\theta = 3.50 \times 10^{-3} \, \text{radians}\)

### Step 3: Calculate the new angular separation
We need to find the new angular separation that is 10% greater than the original:
\[ \text{New } \theta = \theta + 0.1 \theta = 1.1 \theta \]
Calculating this gives:
\[ \text{New } \theta = 1.1 \times (3.50 \times 10^{-3}) = 3.85 \times 10^{-3} \, \text{radians} \]

### Step 4: Set up the equation for the new wavelength
We know the distance \(d\) remains constant, so we can set up the equation for the new wavelength \(\lambda_{new}\):
\[ \text{New } \theta = \frac{\lambda_{new}}{d} \]
Substituting the new angular separation:
\[ 3.85 \times 10^{-3} = \frac{\lambda_{new}}{d} \]

### Step 5: Relate the new wavelength to the original wavelength
From the original equation, we can express \(d\) in terms of the original wavelength:
\[ d = \frac{\lambda}{\theta} = \frac{589 \times 10^{-9}}{3.50 \times 10^{-3}} \]
Now substituting \(d\) into the equation for the new wavelength:
\[ 3.85 \times 10^{-3} = \frac{\lambda_{new}}{\frac{589 \times 10^{-9}}{3.50 \times 10^{-3}}} \]

### Step 6: Solve for \(\lambda_{new}\)
Cross-multiplying gives:
\[ \lambda_{new} = 3.85 \times 10^{-3} \times \frac{589 \times 10^{-9}}{3.50 \times 10^{-3}} \]
Calculating this:
\[ \lambda_{new} = \frac{3.85 \times 589 \times 10^{-9}}{3.50} = \frac{2266.65 \times 10^{-9}}{3.50} \approx 648.0 \times 10^{-9} \, \text{m} \]
Converting this back to nanometers:
\[ \lambda_{new} \approx 648.0 \, \text{nm} \]

### Final Answer
The new wavelength for which the angular separation is 10% greater is approximately **648 nm**.