Question

In a double slit experiment the angular width of a fringe is found to be $0.2^o$ , find the angular width of the fringe if entire experimental apparatus is immersed in a liquid with refractive index $\frac{4}{3}$ .

• A. $0.15^o$
• B. $1^o$
• C. $2^o$
• D. $0.3^o$

Answer 1

Answer: (A)

Since a fringe of width $\beta$ is formed on the screen.
At distance $D$ from the slits, so the angular fringe width
$\theta =\frac{\beta }{D}=\frac{D\lambda /d}{D}=\frac{\lambda }{d}\left[\therefore \beta =\frac{D\lambda }{d}\right]$
$⟹d=\frac{\lambda }{\theta }$
If the wavelength in water be ${\lambda }^{\prime }$ and the angular fringe
Width be ${\theta }^{{}^{\prime }},$ then
$d=\frac{{\lambda }^{\prime }}{{\theta }^{{}^{\prime }}}$ or $\frac{\lambda }{\theta }=\frac{{\lambda }^{\prime }}{{\theta }^{\prime }}$ or ${\theta }^{{}^{\prime }}=\frac{{\lambda }^{\prime }}{\lambda }.{\theta }^{{}^{\prime }}=\frac{\lambda /\mu }{\lambda }.\theta \left[\therefore {\lambda }^{{}^{\prime }}=\frac{\lambda }{\mu }\right]$
$\therefore {\theta }^{{}^{\prime }}=\frac{0.2^o}{4/3}=0.15^o$