Question:
In Young's double slit experiment, the slits are \(3\,\text{mm}\) apart and are illuminated by light of two wavelengths \(3750\,\text{\AA}\) and \(7500\,\text{\AA}\). The screen is placed at \(4\,\text{m}\) from the slits. The minimum distance from the common central bright fringe on the screen at which the bright fringe of one interference pattern due to one wavelength coincides with the bright fringe of the other is
In Young's double slit experiment, the slits are \(3\,\text{mm}\) apart and are illuminated by light of two wavelengths \(3750\,\text{\AA}\) and \(7500\,\text{\AA}\). The screen is placed at \(4\,\text{m}\) from the slits. The minimum distance from the common central bright fringe on the screen at which the bright fringe of one interference pattern due to one wavelength coincides with the bright fringe of the other is
Show Hint
For coincident bright fringes in Young's double slit experiment, use
\[
n_1\lambda_1=n_2\lambda_2
\]
and choose the smallest positive integer values of \(n_1\) and \(n_2\) to find the minimum non-zero distance.
Updated On: Jun 22, 2026
- \(2\,\text{mm}\)
- \(3\,\text{mm}\)
- \(1\,\text{mm}\)
- \(8\,\text{mm}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the condition for coincidence of bright fringes.
In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe is given by \[ y_n=\frac{n\lambda D}{d} \] For two wavelengths, bright fringes coincide when \[ n_1\lambda_1=n_2\lambda_2 \]
Step 2: Substitute the given wavelengths.
Given, \[ \lambda_1=3750\,\text{\AA} \] \[ \lambda_2=7500\,\text{\AA} \] Now, \[ n_1\lambda_1=n_2\lambda_2 \] \[ n_1(3750)=n_2(7500) \] \[ n_1=2n_2 \] For the minimum non-zero distance from the central bright fringe, take the smallest positive integers: \[ n_2=1 \] \[ n_1=2 \] So, the first coincidence occurs when the second bright fringe of wavelength \(3750\,\text{\AA}\) coincides with the first bright fringe of wavelength \(7500\,\text{\AA}\).
Step 3: Convert the given quantities into SI units.
The slit separation is \[ d=3\,\text{mm}=3\times 10^{-3}\,\text{m} \] The screen distance is \[ D=4\,\text{m} \] The longer wavelength is \[ \lambda_2=7500\,\text{\AA} \] Since, \[ 1\,\text{\AA}=10^{-10}\,\text{m} \] we get \[ \lambda_2=7500\times 10^{-10}\,\text{m} \] \[ \lambda_2=7.5\times 10^{-7}\,\text{m} \]
Step 4: Find the distance of the first coincident bright fringe.
Using \[ y=\frac{n_2\lambda_2D}{d} \] Since \(n_2=1\), \[ y=\frac{(1)(7.5\times 10^{-7})(4)} {3\times 10^{-3}} \] \[ y=\frac{30\times 10^{-7}} {3\times 10^{-3}} \] \[ y=10\times 10^{-4}\,\text{m} \] \[ y=10^{-3}\,\text{m} \] \[ y=1\,\text{mm} \]
Step 5: Final conclusion.
Therefore, the minimum distance from the common central bright fringe is \[ \boxed{1\,\text{mm}} \]
In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe is given by \[ y_n=\frac{n\lambda D}{d} \] For two wavelengths, bright fringes coincide when \[ n_1\lambda_1=n_2\lambda_2 \]
Step 2: Substitute the given wavelengths.
Given, \[ \lambda_1=3750\,\text{\AA} \] \[ \lambda_2=7500\,\text{\AA} \] Now, \[ n_1\lambda_1=n_2\lambda_2 \] \[ n_1(3750)=n_2(7500) \] \[ n_1=2n_2 \] For the minimum non-zero distance from the central bright fringe, take the smallest positive integers: \[ n_2=1 \] \[ n_1=2 \] So, the first coincidence occurs when the second bright fringe of wavelength \(3750\,\text{\AA}\) coincides with the first bright fringe of wavelength \(7500\,\text{\AA}\).
Step 3: Convert the given quantities into SI units.
The slit separation is \[ d=3\,\text{mm}=3\times 10^{-3}\,\text{m} \] The screen distance is \[ D=4\,\text{m} \] The longer wavelength is \[ \lambda_2=7500\,\text{\AA} \] Since, \[ 1\,\text{\AA}=10^{-10}\,\text{m} \] we get \[ \lambda_2=7500\times 10^{-10}\,\text{m} \] \[ \lambda_2=7.5\times 10^{-7}\,\text{m} \]
Step 4: Find the distance of the first coincident bright fringe.
Using \[ y=\frac{n_2\lambda_2D}{d} \] Since \(n_2=1\), \[ y=\frac{(1)(7.5\times 10^{-7})(4)} {3\times 10^{-3}} \] \[ y=\frac{30\times 10^{-7}} {3\times 10^{-3}} \] \[ y=10\times 10^{-4}\,\text{m} \] \[ y=10^{-3}\,\text{m} \] \[ y=1\,\text{mm} \]
Step 5: Final conclusion.
Therefore, the minimum distance from the common central bright fringe is \[ \boxed{1\,\text{mm}} \]
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