Question:
In a Young's double slit experiment, the slits are separated by \(0.28\,\text{mm}\) and the screen is placed \(1.4\,\text{m}\) away from the slits. The distance between the central bright fringe and the \(4^{\text{th}}\) order bright fringe is measured to be \(1.2\,\text{cm}\). The wavelength of light used in this experiment is
In a Young's double slit experiment, the slits are separated by \(0.28\,\text{mm}\) and the screen is placed \(1.4\,\text{m}\) away from the slits. The distance between the central bright fringe and the \(4^{\text{th}}\) order bright fringe is measured to be \(1.2\,\text{cm}\). The wavelength of light used in this experiment is
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In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe is
\[
y_n=\frac{n\lambda D}{d}.
\]
Always convert all given quantities into SI units before substitution.
Updated On: Jun 18, 2026
- \(2400\,\text{nm}\)
- \(600\,\text{nm}\)
- \(1200\,\text{nm}\)
- \(800\,\text{nm}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for position of bright fringe.
In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe from the central bright fringe is \[ y_n=\frac{n\lambda D}{d} \] where \[ y_n=\text{distance of }n^{\text{th}}\text{ bright fringe from central bright fringe}, \] \[ n=\text{order of bright fringe}, \] \[ \lambda=\text{wavelength of light}, \] \[ D=\text{distance between slits and screen}, \] and \[ d=\text{separation between slits}. \]
Step 2: Write the given values.
Given, \[ d=0.28\,\text{mm}=0.28\times 10^{-3}\,\text{m} \] \[ D=1.4\,\text{m} \] \[ y_4=1.2\,\text{cm}=1.2\times 10^{-2}\,\text{m} \] \[ n=4 \]
Step 3: Substitute in the formula.
From \[ y_n=\frac{n\lambda D}{d}, \] we get \[ \lambda=\frac{y_n d}{nD} \] Substituting the values, \[ \lambda= \frac{(1.2\times 10^{-2})(0.28\times 10^{-3})}{4\times 1.4} \] \[ \lambda= \frac{0.336\times 10^{-5}}{5.6} \] \[ \lambda=0.06\times 10^{-5}\,\text{m} \] \[ \lambda=6\times 10^{-7}\,\text{m} \]
Step 4: Convert metre into nanometre.
We know, \[ 1\,\text{nm}=10^{-9}\,\text{m} \] Therefore, \[ 6\times 10^{-7}\,\text{m} = 600\times 10^{-9}\,\text{m} \] \[ =600\,\text{nm} \]
Step 5: Final conclusion.
Hence, the wavelength of light used is \[ \boxed{600\,\text{nm}} \]
In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe from the central bright fringe is \[ y_n=\frac{n\lambda D}{d} \] where \[ y_n=\text{distance of }n^{\text{th}}\text{ bright fringe from central bright fringe}, \] \[ n=\text{order of bright fringe}, \] \[ \lambda=\text{wavelength of light}, \] \[ D=\text{distance between slits and screen}, \] and \[ d=\text{separation between slits}. \]
Step 2: Write the given values.
Given, \[ d=0.28\,\text{mm}=0.28\times 10^{-3}\,\text{m} \] \[ D=1.4\,\text{m} \] \[ y_4=1.2\,\text{cm}=1.2\times 10^{-2}\,\text{m} \] \[ n=4 \]
Step 3: Substitute in the formula.
From \[ y_n=\frac{n\lambda D}{d}, \] we get \[ \lambda=\frac{y_n d}{nD} \] Substituting the values, \[ \lambda= \frac{(1.2\times 10^{-2})(0.28\times 10^{-3})}{4\times 1.4} \] \[ \lambda= \frac{0.336\times 10^{-5}}{5.6} \] \[ \lambda=0.06\times 10^{-5}\,\text{m} \] \[ \lambda=6\times 10^{-7}\,\text{m} \]
Step 4: Convert metre into nanometre.
We know, \[ 1\,\text{nm}=10^{-9}\,\text{m} \] Therefore, \[ 6\times 10^{-7}\,\text{m} = 600\times 10^{-9}\,\text{m} \] \[ =600\,\text{nm} \]
Step 5: Final conclusion.
Hence, the wavelength of light used is \[ \boxed{600\,\text{nm}} \]
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