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 defined as: \[ \beta = \frac{\lambda D}{d} \] where:
\( \lambda \) represents the wavelength of light,
\( D \) is the separation between the slits and the screen,
\( d \) is the separation between the two slits.
The fringe width quantifies the distance between adjacent bright fringes. 
Step 2: {Using the Provided Data} 
The separation between the first and second-order bright fringes is given as: \[ y_2 - y_1 = 0.553 { mm} = 0.553 \times 10^{-3} { m} \] For the first bright fringe, its position is: \[ y_1 = \frac{\lambda D}{d} \] For the second bright fringe, its position is: \[ y_2 = \frac{2\lambda D}{d} \] Consequently, the difference in positions is: \[ y_2 - y_1 = \frac{2\lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d} \] 
Step 3: {Calculating \( \lambda \)} 
The wavelength \( \lambda \) can be isolated as: \[ \lambda = \frac{(y_2 - y_1) \cdot d}{D} \] Substituting the provided values: \[ \lambda = \frac{(0.553 \times 10^{-3}) \times (2.08 \times 10^{-3})}{1.8} \] \[ \lambda = \frac{1.15024 \times 10^{-6}}{1.8} \] \[ \lambda = 639 \times 10^{-9} { m} = 639 { nm} \] The calculated wavelength is \( 639 \) nm.