Question

In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index $\frac{4}{3},$ without disturbing the geometrical arrangement, the new fringe width will be

• A. 0.30 mm
• B. 0.40 mm
• C. 0.53 mm
• D. 450 $\mu m$

Answer 1

Answer: (A)

In Young's double slit experiment if distance between coherent source is d, and distance between screen and source is D, then fringe width $\beta$ is given by
$\beta =\frac{D\lambda }{d}....(i)$
where $\lambda$ is wavelength of monochromatic light source.
Also, from Huygen's principle
$\mu =\frac{\lambda }{{\lambda }_w}=$ refractive index
ln water
${\beta }^{\prime }=\frac{D{\lambda }_w}{d}....(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{\beta }{{\beta }^{\prime }}=\frac{\lambda }{{\lambda }_w}\Rightarrow {\beta }^{\prime }=\frac{{\lambda }_w}{\lambda }\beta$
$=\frac{\beta }{\mu }$
Given $\beta =0.4mm,\mu =\frac{4}{3},$ we have
${\beta }^{\prime }=\frac{0.4}{(4/3)}=0.3mm$