Discussion
In Young's double slit experiment performed in air medium, the fringe width observed is 1.4 mm. If the entire arrangement is kept in a liquid medium of refractive index 1.4, then the fringe width (in mm) will be
Answer: B
The fringe width becomes 1.0 mm.
Fringe width \(\beta = \frac{\lambda D}{d}\). In a medium of refractive index \(\mu\) the wavelength becomes \(\lambda/\mu\), so
\[ \begin{aligned} \beta' &= \frac{\beta}{\mu} = \frac{1.4}{1.4} \\ &= 1.0\ \text{mm} \end{aligned} \]
1.4 mm ignores the change of wavelength, 0.7 mm divides by 2 instead of 1.4, and 2.8 mm multiplies instead of dividing.
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