The Young's double slit experiment, angular width of fringes is 0.20 radians for sodium light of wavelength…
The Young's double slit experiment, angular width of fringes is 0.20 radians for sodium light of wavelength $5890 Å$. If the complete system is dipped in water, then the angular width of fringes will be
$\text { [Refractive index of water } \left.=\frac{4}{3}\right]$
$0.30^{\circ}$
$0.15^{\circ}$
$0.11^{\circ}$
$0.22^{\circ}$
Solution
Angular fringe width is given by
$\beta=\frac{\lambda}{d}$
$\beta=0.2$ radians $=\frac{\lambda}{d}$
If dipped in water
$\beta^{\prime}=\frac{\lambda^{\prime}}{d}$
where $\lambda^{\prime}=\frac{\lambda}{\mu}$ for water, $\mu=\frac{4}{3}$
Taking ratio of $\mathrm{eq}^{\mathrm{n}}$ (1) and $\mathrm{eq}^{\mathrm{n}}$ (2)
$\begin{aligned}
& \frac{0.2}{\beta^{\prime}}=\frac{\left(\frac{\lambda}{d}\right)}{\left(\frac{3 \lambda}{4 d}\right)} \\
& \Rightarrow \beta^{\prime}=0.2 \times \frac{3}{4} \text { radians }=0.15 \text { radians }
\end{aligned}$
Taking ratio of $\mathrm{eq}^{\mathrm{n}}$ (1) and $\mathrm{eq}^{\mathrm{n}}$ (2)
$\begin{aligned}
& \frac{0.2}{\beta^{\prime}}=\frac{\left(\frac{\lambda}{d}\right)}{\left(\frac{3 \lambda}{4 d}\right)} \\
& \Rightarrow \beta^{\prime}=0.2 \times \frac{3}{4} \text { radians }=0.15 \text { radians }
\end{aligned}$