Question

A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light (λ=700 nm in vacuum). Find the fringe-width of the pattern formed on the screen.

Answer 1

The Fringe width of the pattern formed on the screen = 0.9 mm

D = distance of the screen from the slits = 0.48 m

d = separation between the slits = 0.28 mm = 0.28 × 10^-3 m

λa = wavelength of red light in vacuum = 700 nm = 700 × 10^-9 m

We must find the wavelength of red light in water(λw) inorder to find the Fringe width .

Refractive index of water = 4/3 = λa/λw

=> λw = λa × 3/4 = 525 × 10^-9 m

Fringe width = (λw × D)/d

=> (525 × 10^-9 × 0.48)/0.28 × 10^-3

=> 9 × 10^-4 m

Fringe width of the pattern = 0.9 mm

Answer 2

The fringe-width of the pattern formed on the screen is $0.90 \mathrm{mm}$

Explanation:

Divide between two d slits $=0.28 \mathrm{mm}=0.28 \times 10^{-3} \mathrm{m}$

Gap between screen and slit $(D)=48 \mathrm{cm}=0.48 \mathrm{m}$

Red Light Wavelength, $\lambda \mathrm{a}=700 \mathrm{nm}$ in vaccum $=700 \times 10^{-9} \mathrm{m}$

Let the red light be wavelength in water $=\lambda \omega$

We realized the Water refractive index $\left(\mu w=\frac{4}{3}\right)$

$\mu W=$ Speed of light in vacuum / Speed of light in the water

$\mu w=\frac{v a}{v \omega}=\frac{\lambda a}{\lambda \omega}$

$\frac{4}{3}=\frac{\lambda a}{\lambda \omega}$

$\Rightarrow \lambda \omega=\frac{3 \lambda a}{4}=3 \times \frac{700}{4}$

$=525 \mathrm{nm}$

So, the pattern fringe width is given by

$\beta=\frac{\lambda \omega D}{d}$

$=\frac{525 \times 10^{-9} \times 0.48}{0.28 \times 10^{-3}}$

$=9 \times 10^{-4}$

$=0.90 \mathrm{mm}$

Fringe-width of the pattern shaped on the screen therefore is $0.90 \mathrm{mm}$.