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 $\lambda_1: \lambda_2=3: 2$