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' = \frac{D \lambda_w}{d} ....( ii)$
Dividing Eq. (i) by Eq. (ii), we get
$ \frac{\beta}{\beta'} = \frac{\lambda}{\lambda_w} \Rightarrow \, \, \beta' = \frac{\lambda_w}{\lambda} \beta$
$ = \frac{\beta}{\mu}$
Given $\beta = 0.4 \, mm, \, \, \mu = \frac{4}{3},$ we have
$ \beta' = \frac{0.4}{(4/3)} = 0.3 \, mm$