Question

In young's double slit experiment the two slits are illuminated by light of wavelength $5890\dot{A}$ and it is observed that successive maxima or minima are seperated by angular displacement $2^{\circ }$. The whole apparatus is immersed in water, then find out angular fringe width is $({\mu }_w=\frac{4}{3})$.

• A. ${0.15}^{\circ }$
• B. ${0.30}^{\circ }$
• C. ${0.5}^{\circ }$
• D. ${0.75}^{\circ }$

Answer 1

The correct option is A ${0.15}^{\circ }$

According to question, the angular fringe width in air is :
${\alpha }_{air}=\frac{{\lambda }_a}{d}={0.2}^{\circ }$

Similarly, angular fringe width in water
${\alpha }_{water}=\frac{{\lambda }_w}{d}$

$\Rightarrow \frac{{\alpha }_{air}}{{\alpha }_{water}}=\frac{{\lambda }_a}{{\lambda }_w}$

$\Rightarrow {\alpha }_{water}=\frac{{\alpha }_{air}{\lambda }_w}{{\lambda }_a}$

By snell's law, $\frac{{\lambda }_a}{{\lambda }_w}={\mu }_w$

$\Rightarrow {\alpha }_w=\frac{{\alpha }_{air}}{{\mu }_w}=\frac{0.2}{\left(\frac{4}{3}\right)}$
$\Rightarrow {\alpha }_w={0.15}^{\circ }$

Hence, option $\left(A\right)$ is correct.