Question

In Young’s double slit experiment, the screen is placed 1m away from the coherent sources. If the wavelength of light used changes from 500 nm to 600 nm the fringe width increases by 0.25mm. The distance between the slits in mm is

• A. 4
• B. 2
• C. 0.2
• D. 0.4

Hint:

Always be careful with unit conversions (nm to m, and m back to mm). A common error is a factor of 10 discrepancy in the final decimal placement.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Fringe width (\(\beta\)) in Young's Double Slit Experiment is the distance between two consecutive bright or dark fringes.
It is directly proportional to the wavelength of light used.

Step 2: Key Formula or Approach:
Fringe width \(\beta = \frac{\lambda D}{d}\)
Change in fringe width \(\Delta \beta = \frac{D}{d}(\lambda_2 - \lambda_1)\)

Step 3: Detailed Explanation:
Given:
\(D = 1\) m
\(\lambda_1 = 500 \text{ nm} = 500 \times 10^{-9}\) m
\(\lambda_2 = 600 \text{ nm} = 600 \times 10^{-9}\) m
\(\Delta \beta = 0.25 \text{ mm} = 0.25 \times 10^{-3}\) m
Using the formula:
\[ 0.25 \times 10^{-3} = \frac{1}{d} \times (600 - 500) \times 10^{-9} \]
\[ 0.25 \times 10^{-3} = \frac{100 \times 10^{-9}}{d} \]
\[ d = \frac{10^{-7}}{0.25 \times 10^{-3}} = \frac{10^{-4}}{0.25} \]
\[ d = 4 \times 10^{-4} \text{ m} \]
Convert to mm:
\[ d = 0.4 \text{ mm} \]

Step 4: Final Answer:
The distance between the slits is 0.4 mm.