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: Understand the Concept:
This problem involves a direct application of the fringe width formula in Young's double-slit experiment (YDSE). Fringe width denotes the separation between adjacent bright or dark fringes.

Step 2: Identify the Formula/Approach:
The fringe width (\(\beta\)) in YDSE is calculated using the formula:\[ \beta = \frac{\lambda D}{d} \]Here, \(\lambda\) is the light's wavelength, \(D\) is the slit-to-screen distance, and \(d\) is the distance between the slits.

Step 3: Detailed Calculation:
Given:
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:
Ensure all units are consistent (preferably SI units).Substitute the given values into the formula:\[ \beta = \frac{(6 \times 10^{-7} \, \text{m}) \times (1.2 \, \text{m})}{1.5 \times 10^{-3} \, \text{m}} \]\[ \beta = \frac{7.2 \times 10^{-7}}{1.5 \times 10^{-3}} \, \text{m} \]\[ \beta = 4.8 \times 10^{-4} \, \text{m} \]Since the options are in millimeters (mm), convert meters to millimeters by multiplying 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 Result:
The calculated fringe width is 0.48 mm.