Q. In Young's double-slit experiment, $12$ fringes are obtained on a certain segment of the screen when light of wavelength $600 \, nm$ is used. If the wavelength of light is changed to $400 \, nm$ , the number of fringes observed on the same segment of the screen is given by
NTA AbhyasNTA Abhyas 2022
Solution:
If on a certain segment of the screen, $n_{1}$ fringes of wavelength $\lambda _{1}$ are present and $n_{2}$ fringes of wavelength $\lambda _{2}$ are present, then
$n_{1}\beta _{1}=n_{2}\beta _{2}$
Here, $\beta _{1}$ and $\beta _{2}$ are fringe widths due to those waves
$\Rightarrow n_{1}\frac{\lambda _{1} D}{d}=n_{2}\frac{\lambda _{2} D}{d}$
$\Rightarrow n_{1}\lambda _{1}=n_{2}\lambda _{2}$
$\therefore n_{2}=\frac{n_{1} \, \lambda _{1}}{\lambda _{2}}=\frac{12 \times 600}{400}=18$
Hence, there will be $18$ fringes formed on the screen.
$n_{1}\beta _{1}=n_{2}\beta _{2}$
Here, $\beta _{1}$ and $\beta _{2}$ are fringe widths due to those waves
$\Rightarrow n_{1}\frac{\lambda _{1} D}{d}=n_{2}\frac{\lambda _{2} D}{d}$
$\Rightarrow n_{1}\lambda _{1}=n_{2}\lambda _{2}$
$\therefore n_{2}=\frac{n_{1} \, \lambda _{1}}{\lambda _{2}}=\frac{12 \times 600}{400}=18$
Hence, there will be $18$ fringes formed on the screen.