Question

In a Young's double slit experiment with a light having wavelength 500 nm, separation between slit is 0.2 mm, distance between slit and screen is 2m. Find fringe width.

Hint:

To avoid extremely common power-of-ten calculation errors, strictly convert all lengths ($\text{nm}$, $\text{mm}$, $\text{cm}$) into meters before plugging them into the formula.

Answer 1

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), light waves from two coherent sources interfere to create a pattern of alternating bright and dark fringes on a distant screen.
The constant spatial distance between any two consecutive bright or dark fringes is defined as the fringe width.
Step 2: Key Formula or Approach:
The standard expression for calculating the fringe width $\beta$ is:
\[ \beta = \frac{\lambda D}{d} \] Where $\lambda$ represents the wavelength of light, $D$ is the macroscopic distance to the screen, and $d$ is the microscopic distance between the two slits.
Step 3: Detailed Explanation:
First, identify and carefully convert all given values into standard SI units (meters).
Wavelength $\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}$.
Distance to screen $D = 2\text{ m}$.
Slit separation $d = 0.2\text{ mm} = 0.2 \times 10^{-3}\text{ m} = 2 \times 10^{-4}\text{ m}$.
Now, substitute these formatted values directly into the fringe width formula:
\[ \beta = \frac{(500 \times 10^{-9}) \times 2}{2 \times 10^{-4}} \] \[ \beta = \frac{1000 \times 10^{-9}}{2 \times 10^{-4}} \] \[ \beta = \frac{10^{-6}}{2 \times 10^{-4}} \] \[ \beta = 0.5 \times 10^{-2}\text{ m} = 5 \times 10^{-3}\text{ m} \] Converting the final result from meters to millimeters yields $5\text{ mm}$.
Step 4: Final Answer:
The calculated fringe width is $5\text{ mm}$.