Question

In Young's double-slit experiment, two different light beams of wavelengths ${\lambda }_1$ and ${\lambda }_2$ produce interference pattern with band widths ${\beta }_1$ and ${\beta }_2$ respectively. If the ratio between ${\beta }_1$ and ${\beta }_2$ is $3:2$, then the ratio between ${\lambda }_1$ and ${\lambda }_2$ is

• A. $3:1$
• B. $1:3$
• C. $2:3$
• D. $3:2$
• E. $4:5$

Answer 1

Answer: (C)

Given,
$\frac{{\beta }_1}{{\beta }_2}=\frac{3}{2}$
As, fringe width in the case of YDSE is given as
$\beta =\frac{\lambda D}{d}$ [here, $D$ and $d$ are constants ]
where, $\lambda$ is the wavelength of light beam.
$\Rightarrow \beta \propto \lambda$
$\Rightarrow \frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{{\lambda }_2}$
$\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{3}{2}$
or ${\lambda }_1:{\lambda }_2=3:2$