Question

In a Young’s double slit experiment, $12$ fringes are observed to be formed in a certain segment of the screen when light of wave length $600nm$ is used. If the wavelength of light is changed to $400nm$, the number of fringes observed in the same segment of the screen is given by

• A. $12$
• B. $18$
• C. $24$
• D. $30$

Answer 1

The correct option is B $18$

Width of the screen having $m$ fringes is $Y_m=\frac{m\lambda D}{d}$
Wavelength of light used is inversely proportional to number of fringes observed for a given segment on screen.

$m_1{\lambda }_1=m_2{\lambda }_2$
Give $m_1=12$
${\lambda }_1=600nm$
${\lambda }_2=400nm$

$\Rightarrow m_2=\frac{m_1{\lambda }_1}{{\lambda }_2}=\frac{12\times 600}{400}=18$