Question

A young's double slit apparatus has slits separated by 0.25 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ($\lambda $=700 nm in vacuum ). Find the fringe – width of the pattern formed on the screen. (${\mu _w}$=4/3)

A. 1$\mu m$

B. 1m

C. 1 cm

D. 1 mm

Answer 1

Hint: To solve this type of question firstly we have to find all the variables which are useful for finding the fringe width. So we will find the wavelength of red light in water. And all other quantities are given already. After finding all useful quantities we will put all the variables in the fringe width formula and finally we will get the fringe width.

Complete step by step solution:

we all know that the formula for the fringe width Is :

$\beta = \dfrac{{\lambda D}}{d}$

Where $\beta $ Is the fringe width in the medium

$\lambda $ is the wavelength of the light in the medium.

D is the separation between slits and screen.

d is the separation between slits

so, we have given that:

D=48 cm or $4.8 \times {10^{ - 1}}$m

d=0.25 mm or $0.25 \times {10^{ - 3}}$m

now we have to find the wavelength of red light in the water.

$\therefore $ We have ${\mu _w} = \dfrac{4}{3}$ i.e. refractive index of water with respect to air

So, we all know that ${\mu _w} = \dfrac{{{n_w}}}{{{n_a}}}$i.e. $\dfrac{{{\text{refractive index of water}}}}{{refractive{\text{ index of air}}}}$

And also ${\mu _w} = \dfrac{{{\text{Wavelength of light in air}}}}{{{\text{ wavelength of light water}}}} = \dfrac{{{\lambda _a}}}{{{\lambda _w}}}$

On equating the above two equations:

$\dfrac{{{n_w}}}{{{n_a}}} = \dfrac{{{\lambda _a}}}{{{\lambda _w}}}$ , where n is refractive index and $\lambda $ is wavelength of light used

So, putting the values of known variables in the above equation:

$  \dfrac{{\dfrac{4}{3}}}{1} = \dfrac{{700}}{{{\lambda _w}}} $

$ \Rightarrow {\lambda _w} = \dfrac{{700 \times 3}}{4} $

$ \Rightarrow {\lambda _w} = 525{\text{ nm}} $

Now we have all quantities to find the fringe width:

So applying the fringe width equation and putting the values of all variables:

$\beta = \dfrac{{{\lambda _w}D}}{d}$

$\Rightarrow \beta = \dfrac{{525 \times {{10}^{ - 9}} \times 4.8 \times {{10}^{ - 1}}}}{{0.25 \times {{10}^{ - 3}}}} $

$\Rightarrow \beta = \dfrac{{2.52 \times {{10}^{ - 7}}}}{{2.5 \times 10 ^{- 4}}} $

$ \Rightarrow \beta = 1.00 \times {10^{ - 3}}{\text{ or 1 mm}} $

Therefore, the correct answer will be option number (D).

Note:
Refractive index: Refractive index is defined as ratio of wavelength of light in air to wavelength of the light in the medium.

Mathematically, ${\mu _w} = \dfrac{{wavelength{\text{ of light in air}}}}{{wavelength{\text{ of light water}}}} = \dfrac{{{\lambda _a}}}{{{\lambda _w}}}$