Question

In a double slit experiment the angular width of a fringe is found to be ${0.2}^{\circ }$ on a screeen $1\text{m}$ away. The wavelength of light used is $6000\AA$. What will be the angular width of a fringe if the entire experimental arrangement is immersed in water? Refractive index of water $=\frac{4}{3}$.

• A. ${0.15}^{\circ }$
• B. ${0.18}^{\circ }$
• C. ${0.2}^{\circ }$
• D. ${0.27}^{\circ }$

Answer 1

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

The angular width of a fringe is given by
$\theta =\frac{\lambda }{d}$
In air: ${\theta }_a=\frac{{\lambda }_a}{d}$
In water: ${\theta }_w=\frac{{\lambda }_w}{d}$
$\therefore \frac{{\theta }_w}{{\theta }_a}=\frac{{\lambda }_w}{{\lambda }_a}$ or ${\theta }_w=\frac{{\lambda }_w{\theta }_a}{{\lambda }_a}...\left(1\right)$
Now, refractive index is defined as
${\mu }_w=\frac{\text{speed of light in air}}{\text{speed of light in water}}$
$=\frac{v_a}{v_w}=\frac{\nu {\lambda }_a}{\nu {\lambda }_w}=\frac{{\lambda }_a}{{\lambda }_w}...\left(2\right)$
where $\nu$ is the frequency of light which remains unchanged.
Using Eq. $\left(2\right)$ in Eq. $\left(1\right)$ we have
${\theta }_w=\frac{{\theta }_a}{{\mu }_w}=\frac{{0.2}^{\circ }}{4/3}={0.15}^{\circ }$
So, the correct choice is (a).