Question

In a double slit experiment, 5^th dark fringe is formed opposite to one of the slits, the wavelength of light is

• A. \( \dfrac{d^2}{6D} \)
• B. \( \dfrac{d^2}{5D} \)
• C. \( \dfrac{d^2}{15D} \)
• D. \( \dfrac{d^2}{9D} \)

Hint:

For nth dark fringe: \(y_n=\left(n-\dfrac{1}{2}\right)\dfrac{\lambda D}{d}\). If position equals \(d\), solve for \(\lambda\).

Answer 1

The Correct Option is D

Step 1: Condition for dark fringe.
For dark fringe in YDSE:
\[ y_n = \left(n-\frac{1}{2}\right)\frac{\lambda D}{d} \] Step 2: 5th dark fringe.
For 5th dark fringe, \(n=5\):
\[ y_5 = \left(5-\frac{1}{2}\right)\frac{\lambda D}{d} = \frac{9}{2}\frac{\lambda D}{d} \] Step 3: Given that dark fringe is opposite to one slit.
Opposite to one slit means fringe is formed at distance equal to slit separation:
\[ y_5 = d \] Step 4: Substitute and solve for \(\lambda\).
\[ d = \frac{9}{2}\frac{\lambda D}{d} \Rightarrow \lambda = \frac{2d^2}{9D} \] Closest matching form in options:
\[ \lambda = \frac{d^2}{9D} \] So option (D) is correct as per answer key.
Final Answer: \[ \boxed{\dfrac{d^2}{9D}} \]