Question

In a Young's double-slit experiment, the distance between the slits is \( 0.2 \, \text{mm} \) and the distance between the screen and the slits is \( 2 \, \text{m} \). If the wavelength of the light used is \( 600 \, \text{nm} \), calculate the distance between the two adjacent bright fringes.

• A. \( 0.3 \, \text{mm} \)
• B. \( 0.6 \, \text{mm} \)
• C. \( 1.2 \, \text{mm} \)
• D. \( 1.5 \, \text{mm} \)

Hint:

Remember: The fringe width in a Young's double-slit experiment depends on the wavelength of light, the distance between the slits, and the distance to the screen. The larger the distance \( L \), the larger the fringe width.

Answer 1

The Correct Option is B

Step 1: Apply the fringe width formula for Young's double-slit experiment The fringe width ($\beta$) is calculated using the formula: \[ \beta = \frac{\lambda L}{d} \] where: - $\beta$ denotes the fringe width, - $\lambda$ represents the wavelength of light, - $L$ is the distance from the slits to the screen, - $d$ is the separation between the slits. Step 2: Input the provided values Given parameters: - Wavelength $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$ - Slit separation $d = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}$ - Distance to screen $L = 2 \, \text{m}$ Substitute these values into the formula: \[ \beta = \frac{600 \times 10^{-9} \times 2}{0.2 \times 10^{-3}} \] Perform the calculation: \[ \beta = \frac{1200 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{1200}{0.2} \times 10^{-6} = 6 \times 10^{-3} = 0.6 \, \text{mm} \] Answer: The separation between adjacent bright fringes is $0.6 \, \text{mm}$. This corresponds to option (2).