In a young's double slit experiment, the slits are separated by $0.28 \mathrm{~mm}$ and the screen is placed…
In a young's double slit experiment, the slits are separated by $0.28 \mathrm{~mm}$ and the screen is placed $1.4 \mathrm{~m}$ away from the slits. The distance between the central bright fringe and the 4 th order bright fringe is measured to be $1.2 \mathrm{~cm}$. The wavelength of light used in this experiment is
$2400 \mathrm{~nm}$
$600 \mathrm{~nm}$
$1200 \mathrm{~nm}$
$800 \mathrm{~nm}$
Solution
Distance between the slits, $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$
Distance between the slits and the screen, $D=1.4 \mathrm{~m}$
Distance between the central fringe and the fourth $(n=4)$ fringe,
$
u=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}
$
In case of a constructive interference, we have the relation for the distance between the two fringes as:
$
u=n \lambda \frac{D}{d}
$
Where,
$
\begin{aligned}
& n=\text { Order of fringes }=4 \\
& \lambda=\text { Wavelength of light used } \\
& \therefore \lambda=\frac{u d}{n D} \\
& =\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4} \\
& =6 \times 10^{-7} \\
& =600 \mathrm{~nm}
\end{aligned}
$
Hence, the wavelength of the light is $600 \mathrm{~nm}$