Q10E - University Physics with Modern Physics

Question

 If the entire apparatus of Exercise 35.9 (slits, screen, and space in between) is immersed in water, what then is the distance between the second and third dark lines?

Step-by-Step Solution

Answer

The distance between the second and third dark lines of interference pattern when the setup is immersed inside water is 0.625 mm.

Step 1: Given Data

Wavelength emitted:  $500 \mathrm{nm}$ 

Distance between slit and screen: $0.75 \mathrm{m}$ 

Distance between slits: $0.45 \mathrm{mm}$

Refractive index of water $1.333$ 

Step 2: (a) Concept of Young’s double slit Experiment.

The redistribution of the intensity of light is when lights from two coherent sources having the same phase are superimposed on each other. This phenomenon is called interference of light.

The expression of location of dark bands here is,

${\mathbf{y}}_{\mathbf{m}}\mathbf{=}\mathbf{R}\frac{\mathbf{m\lambda }}{\mathbf{d}}$                                                                                                   ...(i)

Here, ${\mathbf{y}}_{\mathbf{m}}$ is the distance of the fringe, $\mathbf{\lambda }$ is the wavelength,  $\mathbf{m}$ is the order of the fringe, $\mathbf{R}$ is the distance between the screen and the slits, and $\mathbf{d}$ is the slit width.

Step 3: (b) Determination of the separation between the two slits when the setup is immersed in water.

The wavelength of the light changes when the setup is immersed in water and this change is given as,

$\mathrm{\lambda }=\frac{{\mathrm{\lambda }}_0}{\mathrm{n}}$

 

Here, ${\mathrm{\lambda }}_0$ is the wavelength in air, ${\mathrm{\lambda }}_{}$is the wavelength in water and $n$ is the refractive index of water. 

For water, $n=1.333$.

Distance between adjacent dark lines is given as,

$$\begin{gathered} ∆y=y_{m+1}-y_m \\ =\frac{R\lambda }{d} \\ =\frac{R{\lambda }_{°}}{dn} \end{gathered}$$

Substitute all the values,

$$\begin{gathered} ∆y=\frac{\left(0.750 \mathrm{m}\right)\left(500\times {10}^{-9}\right)}{\left(0.450\times {10}^{-3} \mathrm{m}\right)\left(1.333\right)} \\ =6.25\times {10}^{-4} \\ =0.625 \mathrm{mm} \end{gathered}$$

Thus, the distance between the fringes is 0.625 mm.