Q9E - University Physics with Modern Physics

Question

Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

Step-by-Step Solution

Answer

The distance between the second and third dark lines of interference pattern is $0.83 \mathrm{mm}$.

Step 1: Given Data

Wavelength emitted:  $500 \mathrm{nm}$ 

Distance between slit and screen: $75.0 \mathrm{cm}$ 

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

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 the location of dark bands here is,

$$\begin{gathered} \mathbf{dsin\theta }\mathbf{=}\left(m+\frac{1}{2}\right)\mathbf{\lambda } \\ \mathbf{sin\theta }\mathbf{=}\frac{\left(m+\frac{1}{2}\right)\mathbf{\lambda }}{\mathbf{d}}\mathbf{,}\mathbf{ }\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$                                                          ...(i)                                        

Here, λ is the wavelength, m is the order of the fringe, $\mathit{\theta }$ is the angular separation of fringe from the center of the central fringe, and d is the sli

Step 3:

The first dark line is for m = 0,

The second dark line is for m = 1,

$$\begin{gathered} \sin {\theta }_1=\frac{3\lambda }{2d} \\ =\frac{3\left(500\times {10}^{-9 }\mathrm{m}\right)}{2\left(0.450\times {10}^{-3 }\mathrm{m}\right)} \\ =1.667 \times {10}^{-3 }\mathrm{rad} \end{gathered}$$

 

The third dark line is for m = 2,

$$\begin{gathered} \sin {\theta }_2=\frac{5\lambda }{2d} \\ =\frac{3\left(500\times {10}^{-9} \mathrm{m}\right)}{2\left(0.450\times {10}^{-3}\right)} \\ =2.778\times {10}^{-3} \end{gathered}$$ 

Each dark line is at a distance from the centre of the central bright, the expression of the distance,

${\mathrm{y}}_{\mathrm{m}}=\mathrm{Rtan\theta }$ 

Where R = 0.850 m is the distance between the screen and the slits.

For small values of${\theta }_{,}$ ${\mathrm{y}}_{\mathrm{m}}={\mathrm{R\theta }}_{\mathrm{m}}$,

$$\begin{gathered} y_1=R{\theta }_1 \\ =\left(0.750 \mathrm{ml}\right)\left(2.778\times {10}^{-3} \mathrm{rad}\right) \\ =1.25\times {10}^{-3} \mathrm{m} \end{gathered}$$

$$\begin{gathered} y_2=R{\theta }_2 \\ =\left(0.750 \mathrm{ml}\right)\left(2.778\times {10}^{-3} \mathrm{rad}\right) \\ =2.08\times {10}^{-3} \mathrm{m} \end{gathered}$$

$$\begin{gathered} ∆y=y_2-y_1 \\ =2.08\times {10}^{-3} m-1.25 \times {10}^{-3} \\ =0.83 \mathrm{mm} \end{gathered}$$

 

Thus, the distance between the two slits $0.83 \mathrm{mm}$.