Question

In Young's double-slit experiment, the slits are separated by 0.28 mm, and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 12 cm. Then, the wavelength of light used in the experiment is …….

Hint:

In Young’s double-slit experiment, the fringe width is given by \( \beta = \frac{\lambda D}{d} \). Using this, the wavelength of light can be determined if the fringe separation and slit parameters are known.

Answer 1

Step 1: Understanding the Fringe Formula 
In Young’s double-slit experiment, the fringe width (\( \beta \)) is given by: \[ \beta = \frac{\lambda D}{d} \] where:
- \( \lambda \) = Wavelength of light (in meters),
- \( D \) = Distance between slits and screen (in meters),
- \( d \) = Distance between the two slits (in meters),
- \( \beta \) = Fringe width (in meters). 
Step 2: Given Data 
- Distance between the slits: \( d = 0.28 \) mm \( = 0.28 \times 10^{-3} \) m,
- Distance to the screen: \( D = 1.4 \) m,
- Distance between central bright fringe and fourth bright fringe: \[ y = 12 { cm} = 0.12 { m} \] Since this corresponds to the fourth bright fringe, we use: \[ y = 4\beta \] 
Step 3: Calculating Fringe Width 
\[ \beta = \frac{y}{4} = \frac{0.12}{4} = 0.03 { m} \] 
Step 4: Calculating Wavelength 
Rearranging the fringe width equation: \[ \lambda = \frac{\beta d}{D} \] Substituting values: \[ \lambda = \frac{(0.03) (0.28 \times 10^{-3})}{1.4} \] \[ \lambda = \frac{8.4 \times 10^{-6}}{1.4} = 6 \times 10^{-6} { m} \] 
Step 5: Conclusion 
Since \( 6 \times 10^{-6} \) m is 6000 nm, the correct answer is 6000 nm.