Question:
In a Young’s double slit experiment, a set of parallel slits with a separation of 0.1 mm is illuminated by light having a wavelength of 620 nm. The interference pattern is observed on a screen 4 m from the slits. The difference in the path lengths of the light waves from each of the slits to the location of a 4th (fourth) order bright fringe on the screen is:
In a Young’s double slit experiment, a set of parallel slits with a separation of 0.1 mm is illuminated by light having a wavelength of 620 nm. The interference pattern is observed on a screen 4 m from the slits. The difference in the path lengths of the light waves from each of the slits to the location of a 4th (fourth) order bright fringe on the screen is:
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In YDSE, bright fringe condition is always \(\Delta x = n\lambda\), independent of slit separation and screen distance.
Updated On: Jul 18, 2026
- 2.48 μm
- 1.24 μm
- 155 nm
- 0
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The Correct Option is A
Solution and Explanation
Step 1: Understanding condition for bright fringes in YDSE.
In Young’s Double Slit Experiment, constructive interference (bright fringe) occurs when path difference is given by: \[ \Delta x = n\lambda \] where \(n\) is the order of bright fringe and \(\lambda\) is wavelength of light.
Step 2: Identifying given values.
Here, \[ n = 4, \quad \lambda = 620 \, \text{nm} \] We note that slit separation and screen distance are not required for path difference calculation, only order and wavelength matter.
Step 3: Applying formula for path difference.
\[ \Delta x = n\lambda = 4 \times 620 \, \text{nm} \]
Step 4: Calculation of numerical value.
\[ \Delta x = 2480 \, \text{nm} \]
Step 5: Unit conversion.
Since, \[ 1 \, \mu m = 1000 \, nm \] \[ 2480 \, nm = 2.48 \, \mu m \]
Step 6: Final conclusion.
Thus, path difference at 4th bright fringe is: \[ \boxed{2.48 \, \mu m} \]
In Young’s Double Slit Experiment, constructive interference (bright fringe) occurs when path difference is given by: \[ \Delta x = n\lambda \] where \(n\) is the order of bright fringe and \(\lambda\) is wavelength of light.
Step 2: Identifying given values.
Here, \[ n = 4, \quad \lambda = 620 \, \text{nm} \] We note that slit separation and screen distance are not required for path difference calculation, only order and wavelength matter.
Step 3: Applying formula for path difference.
\[ \Delta x = n\lambda = 4 \times 620 \, \text{nm} \]
Step 4: Calculation of numerical value.
\[ \Delta x = 2480 \, \text{nm} \]
Step 5: Unit conversion.
Since, \[ 1 \, \mu m = 1000 \, nm \] \[ 2480 \, nm = 2.48 \, \mu m \]
Step 6: Final conclusion.
Thus, path difference at 4th bright fringe is: \[ \boxed{2.48 \, \mu m} \]
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