Question

In YDSE, monochromatic light falls on a screen 1.80 m from two slits separated by 2.08 mm. The first and second order bright fringes are separated by 0.553 mm. The wavelength of light used is:

• A. 520 nm
• B. 639 nm
• C. 715 nm
• D. None of these

Hint:

The fringe width in YDSE increases if the wavelength of light or the screen distance increases, but decreases if the slit separation increases.

Answer 1

The Correct Option is B

Step 1: {Understanding Young’s Double-Slit Experiment (YDSE)} 
In Young’s Double-Slit Experiment, the fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] where:
\( \lambda \) is the wavelength of light,
\( D \) is the distance between the slits and the screen,
\( d \) is the distance between the two slits.
The fringe width is the distance between consecutive bright fringes. 
Step 2: {Using the Given Data} 
It is given that the distance between the first and second-order bright fringes is: \[ y_2 - y_1 = 0.553 { mm} = 0.553 \times 10^{-3} { m} \] For the first bright fringe, \[ y_1 = \frac{\lambda D}{d} \] For the second bright fringe, \[ y_2 = \frac{2\lambda D}{d} \] So, \[ y_2 - y_1 = \frac{2\lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d} \] 
Step 3: {Solving for \( \lambda \)} 
\[ \lambda = \frac{(y_2 - y_1) \cdot d}{D} \] Substituting the given values: \[ \lambda = \frac{(0.553 \times 10^{-3}) \times (2.08 \times 10^{-3})}{1.8} \] \[ \lambda = \frac{1.15024 \times 10^{-6}}{1.8} \] \[ \lambda = 639 \times 10^{-9} { m} = 639 { nm} \] Thus, the correct answer is \( 639 \) nm. 
 

Answer 2

Step 1: Understanding Young’s Double Slit Experiment (YDSE)
In YDSE, the position of the bright fringes on the screen is given by the formula for fringe position: \[ y_n = n \frac{\lambda D}{d} \] Where: - \( y_n \) = distance of the \( n^{\text{th}} \) bright fringe from the center, - \( \lambda \) = wavelength of light, - \( D \) = distance from the slits to the screen, - \( d \) = distance between the slits, - \( n \) = order of the fringe.

Step 2: Given Values
- \( D = 1.80 \, \text{m} = 1800 \, \text{mm} \)
- \( d = 2.08 \, \text{mm} \)
- Distance between first and second order bright fringes: \( y_2 - y_1 = 0.553 \, \text{mm} \)

Step 3: Use the formula for fringe separation between two consecutive bright fringes:
\[ y_2 - y_1 = \left(2 - 1\right) \frac{\lambda D}{d} = \frac{\lambda D}{d} \]

Step 4: Solving for \( \lambda \):
\[ \lambda = \frac{(y_2 - y_1) \cdot d}{D} \] \[ \lambda = \frac{0.553 \times 2.08}{1800} \, \text{mm} = \frac{1.15104}{1800} \, \text{mm} = 6.395 \times 10^{-4} \, \text{mm} = 639.5 \, \text{nm} \]

Step 5: Final Answer
\[ \boxed{\lambda \approx 639 \, \text{nm}} \] So the correct answer is: Option 2: 639 nm