Question

In a Young's double slit experiment, light of wavelength $5900\,\mathring{A}$ is used. When the slits are $2\, mm$ apart, the fringe width is $1.2\, mm$. 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} $
$\Rightarrow \, \beta_{2}=0.8 \,mm$