Question

In a Young double slit experiment, the slits are separated by $2.8\,{\text{mm}}$ and the screen is placed $1.4\,{\text{m}}$ away. The distance between the central bright fringe and the fourth dark fringe is measured to be $1.2\,{\text{mm}}$. Determine the wavelength of the light used in the experiment. Also find the distance between the fifth dark fringe from the second bright fringe.

Answer 1

Use the formulae for the distance of nth bright fringe from the central bright fringe and nth dark fringe from the central bright fringe. Substitute the values in the formula and calculate the wavelength of the light used in the experiment. Then rewrite these two formulae for the fifth dark fringe and second bright fringe and take the difference of these two to determine the required distance between the fringes.

Formulae used:
The distance ${y_B}$ of nth bright fringe from the central bright fringe is
${y_B} = \dfrac{{n\lambda D}}{d}$ …… (1)
Here, $\lambda$ is the wavelength of the light, $D$ is the distance between the screen and slits and $d$ is the separation between two slits.
The distance ${y_D}$ of nth dark fringe from the central bright fringe is
${y_D} = \dfrac{{\left( {2n - 1} \right)}}{2}\dfrac{{\lambda D}}{d}$ …… (2)
Here, $\lambda$ is the wavelength of the light, $D$ is the distance between the screen and slits and $d$ is the separation between two slits.

Complete step by step answer:
We have given that the separation between the slits is $2.8\,{\text{mm}}$ and is the distance between the slit and screen is $1.4\,{\text{m}}$.
$d = 2.8\,{\text{mm}}$
$\Rightarrow D = 1.4\,{\text{m}}$
The distance between the central bright fringe and fourth dark fringe is $1.2\,{\text{mm}}$.
${y_4} = 1.2\,{\text{mm}}$
We have asked to calculate the wavelength of the light used in the experiment.Rewrite equation (2) for the distance of the fourth dark fringe from the central bright fringe.
${y_4} = \dfrac{{\left( {2 \times 4 - 1} \right)}}{2}\dfrac{{\lambda D}}{d}$
$\Rightarrow {y_4} = \dfrac{7}{2}\dfrac{{\lambda D}}{d}$
$\Rightarrow \lambda = \dfrac{{2{y_4}d}}{{7D}}$

Substitute $1.2\,{\text{mm}}$ for ${y_4}$, $2.8\,{\text{mm}}$ for $d$ and $1.4\,{\text{m}}$ for $D$ in the above equation.
$\Rightarrow \lambda = \dfrac{{2\left( {1.2\,{\text{mm}}} \right)\left( {2.8\,{\text{mm}}} \right)}}{{7\left( {1.4\,{\text{m}}} \right)}}$
$\Rightarrow \lambda = \dfrac{{2\left( {1.2 \times {{10}^{ - 3}}\,{\text{m}}} \right)\left( {2.8 \times {{10}^{ - 3}}\,{\text{m}}} \right)}}{{7\left( {1.4\,{\text{m}}} \right)}}$
$\Rightarrow \lambda = 0.6857 \times {10^{ - 6}}\,{\text{m}}$
$\Rightarrow \lambda = 6857 \times {10^{ - 10}}\,{\text{m}}$
$\Rightarrow \lambda = 6857\,\mathop {\text{A}}\limits^{\text{o}}$
Hence, the wavelength of the light used in the experiment is $6857\,\mathop {\text{A}}\limits^{\text{o}}$.

We have also asked to calculate the distance between the fifth dark fringe from the second bright fringe.Rewrite equation (1) for the distance of the fifth dark fringe from the central bright fringe.
${y_5} = \dfrac{{\left( {2 \times 5 - 1} \right)}}{2}\dfrac{{\lambda D}}{d}$
$\Rightarrow {y_5} = \dfrac{9}{2}\dfrac{{\lambda D}}{d}$
Rewrite equation (1) for the distance of the second bright fringe from the central bright fringe.
${y_2} = \dfrac{{2\lambda D}}{d}$
The distance between the fifth dark fringe from the second bright fringe is
$\Delta y = {y_5} - {y_2}$
$\Rightarrow \Delta y = \dfrac{9}{2}\dfrac{{\lambda D}}{d} - \dfrac{{2\lambda D}}{d}$
$\Rightarrow \Delta y = \left( {\dfrac{9}{2} - 2} \right)\dfrac{{\lambda D}}{d}$
$\Rightarrow \Delta y = \left( {\dfrac{{9 - 4}}{2}} \right)\dfrac{{\lambda D}}{d}$
$\Rightarrow \Delta y = \dfrac{{5\lambda D}}{{2d}}$

Substitute $6857\,\mathop {\text{A}}\limits^{\text{o}}$ for $\lambda$, $1.4\,{\text{m}}$ for $D$ and $2.8\,{\text{mm}}$ for $d$ in the above equation.
$\Rightarrow \Delta y = \dfrac{{5\left( {6857\,\mathop {\text{A}}\limits^{\text{o}} } \right)\left( {1.4\,{\text{m}}} \right)}}{{2\left( {2.8\,{\text{mm}}} \right)}}$
$\Rightarrow \Delta y = \dfrac{{5\left( {6857 \times {{10}^{ - 10}}\,{\text{m}}} \right)\left( {1.4\,{\text{m}}} \right)}}{{2\left( {2.8 \times {{10}^{ - 3}}\,{\text{m}}} \right)}}$
$\Rightarrow \Delta y = 17142.5 \times {10^{ - 7}}\,{\text{m}}$
$\Rightarrow \Delta y = 1.71425 \times {10^{ - 3}}\,{\text{m}}$
$\therefore \Delta y = 1.71\,{\text{mm}}$

Hence, the distance between the fifth dark fringe and second bright fringe is $1.71\,{\text{mm}}$.

Note: The students may think that the formula for the nth bright and dark fringe from the central bright fringe is the same. But the students should keep in mind that the formulae for both these distances are different and these formulae should be used correctly. Otherwise, the final answer will be incorrect.