Question

A Young double slit experimental setup is immersed in water of refractive index $1.33$. It has slit separation $1\, mm$ and the distance between slits and screen is $1.33\, m$. If the wavelength of incident light on slits is $6300\, \mathring A$, then the fringe width on screen is

• A. 6.3 mm
• B. 0.63 mm
• C. 0.63 m
• D. 6.3 m

Answer 1

Answer: (B)

When Young's double slit experiment setup is immersed in water then fringe width, $y=\frac{\lambda D}{\mu d}$.
Given, wavelength of incident light on slits, $\lambda=6300$
$\mathring{A}=6300 \times 10^{-10} m$
distance between slit and screen, $D=1.33 \,m$,
and refractive index of water $\mu=1.33$
Hence, the fringe width on the screen,
$y =\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 1 \times 10^{-3}}$
$y =6.3 \times 10^{-4} m $
$=0.63 \,mm$