Question

In Young's double slit experiment, the distance between the slits is 3 mm and the slits are 2m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 nm respectively. The separation on the screen between the \(5^{\text{th}}\) order bright fringes on two interference patterns is

• A. \(2\times 10^{-4}\,\text{m}\)
• B. \(1\times 10^{-4}\,\text{m}\)
• C. \(8\times 10^{-4}\,\text{m}\)
• D. \(4\times 10^{-4}\,\text{m}\)

Hint:

Position of the nth bright fringe is n lambda D / d.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The \(n\)-th bright fringe is at \(y_n = \dfrac{n\lambda D}{d}\) from the central maximum.

Step 2: Find each position:
\(D = 2\) m, \(d = 3\times10^{-3}\) m, \(n = 5\).
\(y_{600} = \dfrac{5\times600\times10^{-9}\times2}{3\times10^{-3}} = 2\times10^{-3}\) m.
\(y_{480} = \dfrac{5\times480\times10^{-9}\times2}{3\times10^{-3}} = 1.6\times10^{-3}\) m.

Step 3: Separation:
\[ 2\times10^{-3} - 1.6\times10^{-3} = 4\times10^{-4}\ \text{m} \]

Step 4: Check:
Option (D). The formula can also be applied to the wavelength difference of 120 nm directly.

Final Answer:
The two 5th fringes are 4e-4 m apart. \[ \boxed{\text{(D) }4\times10^{-4}\ \text{m}} \]