91Ó°ÊÓ

Slit width $a=0.4mm$

Distance between the two minima on either side of the central maximalocalid="1663048702619" $∆y_{1,-1}=1.8mm$

Wavelength of incident light $\mathrm{λ}=450nm$

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance $\mathrm{θ}$of the $m_{th}$ order minima in diffraction pattern produced from a single slit having slit width $a$ is

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ …(¾±)

Here, $\mathrm{λ}$ is the wavelength of the incident light.

Let the distance from the slit to the screen be$D$ . For small angular distances

$$\begin{gathered} \sin \mathrm{θ}≈\tan \mathrm{θ} \\ =\frac{y}{D} \end{gathered}$$

Here $y$ is the distance measured on the screen. Thus, from equation (i) the separation between the first two minima $mÂ\pm 1$ on either sides of the central maxima is

$$\begin{gathered} \frac{a∆y_{1,-1}}{D}=\left\{1-\left(-1\right)\right\}\mathrm{λ} \\ D=\frac{a∆y_{1,-1}}{2\mathrm{λ}} \end{gathered}$$

Substitute the values to get

$$\begin{gathered} D=\frac{0.4mm×1.8mm}{2×450nm} \\ =\frac{0.4×{10}^{-3}m×1.8×{10}^{-3}m}{2×450×{10}^{-9}m} \\ =0.8m \end{gathered}$$

Thus, the distance is$0.8m$ .

From equation (i), the distance between the first $\left(m=1\right)$ and third minima$\left(m=3\right)$ is

$$\begin{gathered} \frac{a∆y_{1,3}}{D}=\left\{3-1\right\}\mathrm{λ} \\ ∆y_{1,3}=\frac{2\mathrm{λ}D}{a} \\ =\frac{2×450×{10}^{-9}m×0.8m}{0.4×{10}^{-3}m} \\ =1.8mm \end{gathered}$$

Thus, the distance is $1.8mm$.