Question

In an interference pattern of two waves fringe width is $\beta$. If the frequency of source is doubled then fringe width will become:-

• A. $\frac{1}{2}\beta$
• B. $\beta$
• C. $2\beta$
• D. $\frac{3}{2}\beta$

Answer 1

The correct option is A $\frac{1}{2}\beta$

Fringe width is given by,

$\beta =\frac{\lambda D}{d}$

Where, $\lambda$ is the wavelength of light, $D$ is the distance from slit to screen and $d$ is the slit width.

We know,

$c=\lambda v$

$c$ is speed of light and $\nu$ is frequency of light.

If $\nu$ is doubled, then,

$c={\lambda }^{\prime }2v=\lambda v$

$⟹{\lambda }^{\prime }=\frac{\lambda }{2}$

So, the fringe width becomes,

${\beta }^{\prime }=\frac{{\lambda }^{\prime }D}{d}=\frac{1}{2}\frac{\lambda D}{d}=\frac{\beta }{2}$

Thus, the fringe width becomes half of the original value when the frequency of light is doubled.