Question

In a Young's double-slit experiment, two slits are 1.5 mm apart while the screen is 1.2 m away. When a light of wavelength 600 nm is incident on slits, the fringe width will be

• A. 0.48 mm
• B. 4.5 mm
• C. 4.8 mm
• D. 4.2 mm

Hint:

In YDSE calculations, always convert all lengths (\(d, D, \lambda\)) to the same unit, preferably the base SI unit (meters), before plugging them into the formula. This minimizes the chance of unit conversion errors. Convert the final answer to the unit required by the options.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires the direct application of the formula for fringe width in a Young's double-slit experiment (YDSE). The fringe width is the separation between consecutive bright fringes or consecutive dark fringes.

Step 2: Key Formula or Approach:
The fringe width (\(\beta\)) in a YDSE is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where \(\lambda\) is the wavelength of the light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the two slits.

Step 3: Detailed Explanation:
Given data:
Slit separation, \(d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}\).
Screen distance, \(D = 1.2 \, \text{m}\).
Wavelength, \(\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} = 6 \times 10^{-7} \, \text{m}\).
Calculation:
It is essential to use consistent units (SI units are recommended). Substitute the values into the formula: \[ \beta = \frac{(6 \times 10^{-7} \, \text{m}) \times (1.2 \, \text{m})}{1.5 \times 10^{-3} \, \text{m}} \] \[ \beta = \frac{7.2 \times 10^{-7}}{1.5 \times 10^{-3}} \, \text{m} \] \[ \beta = 4.8 \times 10^{-4} \, \text{m} \] The options are in millimeters (mm). To convert from meters to millimeters, multiply by \(10^3\). \[ \beta = (4.8 \times 10^{-4}) \times 10^3 \, \text{mm} = 4.8 \times 10^{-1} \, \text{mm} = 0.48 \, \text{mm} \]

Step 4: Final Answer:
The fringe width will be 0.48 mm.