Monochromatic light (wavelength$\mathbf{=}\mathbf{450}\mathbf{}\mathit{n}\mathit{m}$) is incident perpendicularly on a single slit (width$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}\mathit{m}$). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is $\mathbf{1}\mathbf{.}\mathbf{8}\mathit{m}\mathit{m}$.

(a) What is the distance from the slit to the screen? (Hint:The angle to either minimum is small enough that$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\approx }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$.)

(b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?

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 $\lambda =450nm$

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 $\theta$of the $m_{th}$ order minima in diffraction pattern produced from a single slit having slit width $a$ is

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …(i)

Here, $\lambda$ is the wavelength of the incident light.

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

$$\begin{gathered} \sin \theta \approx \tan \theta \\ =\frac{y}{D} \end{gathered}$$

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

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

Substitute the values to get

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

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

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

Thus, the distance is $1.8mm$.