Step1. Path Difference Condition for Minima: - For a dark fringe (minima) at P, the path difference should be:
\[ 2\sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2} \]
- Rearrange:
\[ \sqrt{D^2 + d^2} - D = \frac{\lambda}{4} \]
- Therefore:
\[ \sqrt{D^2 + d^2} = D + \frac{\lambda}{4} \]
Step 2. Square Both Sides:
\[ D^2 + d^2 = D^2 + D\lambda + \frac{\lambda^2}{16} \]
- Simplify to solve for d:
\[ d^2 = D\lambda + \frac{\lambda^2}{16} \]
Step 3. Substitute Given Values: - \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\), \(D = 0.2 \, \text{m}\)
\[ d^2 = 0.2 \times 400 \times 10^{-9} + \frac{(400 \times 10^{-9})^2}{16} \]
- Calculate:
\[ d^2 \approx 4 \times 10^{-8} \, \text{m}^2 \]
\[ d \approx 2 \times 10^{-4} \, \text{m} = 0.20 \, \text{mm} \]
So, the correct answer is: \(d = 0.20 \, \text{mm}\)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32