91Ó°ÊÓ

Wavelength emitted: $500\mathrm{nm}$

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

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

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,

$$\begin{gathered} \mathbf{»å²õ¾\pm ²Ôθ}\mathbf{=}\left(m+\frac{1}{2}\right)\mathbf{λ} \\ \mathbf{²õ¾\pm ²Ôθ}\mathbf{=}\frac{\left(m+\frac{1}{2}\right)\mathbf{λ}}{\mathbf{d}}\mathbf{,}\mathbf{}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{Â\pm }\mathbf{1}\mathbf{,}\mathbf{Â\pm }\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$ ...(i)

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

The first dark line is for m = 0,

The second dark line is for m = 1,

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

The third dark line is for m = 2,

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

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

${\mathrm{y}}_{\mathrm{m}}=\mathrm{¸é³Ù²¹²Ôθ}$

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

For small values of${\mathrm{θ}}_{,}$${\mathrm{y}}_{\mathrm{m}}={\mathrm{¸éθ}}_{\mathrm{m}}$,

role="math" localid="1663936712432" $$\begin{gathered} y_1=R{\mathrm{θ}}_1 \\ =\left(0.750\mathrm{ml}\right)\left(2.778×{10}^{-3}\mathrm{rad}\right) \\ =1.25×{10}^{-3}\mathrm{m} \end{gathered}$$

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

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

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