Question

In Young's double slit experiment. the fringe width with light of wavelength 6000 $\stackrel{0}{A}$ is 3 mm . The fringe width, when the wavelength of light is changed to 4000 $\stackrel{0}{A}$ is

• A. 3 mm
• B. 1 mm
• C. 2 mm
• D. 4 mm

Answer 1

The correct option is C 2 mm

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

where $\lambda$ is the wave; length of light, D is distance between slits and the screen, d is the distance between the two slits.

As D and d remain the same, $\beta \propto \lambda$

or $\frac{{\beta }^{\prime }}{\beta }=\frac{{\lambda }^{\prime }}{\lambda }or{\beta }^{\prime }=\frac{{\lambda }^{\prime }\beta }{\lambda }$

Subsitituting the given values, we get

${\beta }^{\prime }=\frac{4000\stackrel{0}{A}\times 3mm}{6000\stackrel{0}{A}}=2mm$