Question

In Young's double-slit experiment, $12$ fringes are obtained on a certain segment of the screen when light of wavelength $600nm$ is used. If the wavelength of light is changed to $400nm$ , the number of fringes observed on the same segment of the screen is given by

• A. $18$
• B. $24$
• C. $30$
• D. $36$

Answer 1

Answer: (A)

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.