Question

In a Young's double slit experiment, light of wavelength $5900\mathring{A}$ is used. When the slits are $2mm$ apart, the fringe width is $1.2mm$. If the slit separation is increased to one and half times the previous value, then the fringe width will be

• A. 0.9 mm
• B. 0.8 mm
• C. 1.8 mm
• D. 1.6 mm

Answer 1

Answer: (B)

In a Young's double slit experiment, wavelength of light, $\lambda =5900\mathring{A}$ and distance between the slits, $d_1=d=2\times 10^{-3}m$,
Fringe width, ${\beta }_1=1.2\times 10^{-3}m$ and $d_2=\frac{3}{2}d$
As, the fringe width,
$\beta =\frac{\lambda D}{d}$
So, ${\beta }_1=\frac{{\lambda }_1{D}_1}{d_1}$ and ${\beta }_2=\frac{{\lambda }_2{D}_2}{d_2}$
Given, ${\lambda }_1={\lambda }_2$ and $D_1=D_2$
$\Rightarrow \frac{{\beta }_1}{{\beta }_2}=\frac{d_2}{d_1}$
Putting the given values, we get
$\Rightarrow \frac{1.2\times 10^{-3}}{{\beta }_2}=\frac{\frac{3}{2}d}{d}$
$\Rightarrow {\beta }_2=0.8mm$