# 物理代写|力学代写mechanics代考|FNEG1004

## 物理代写|力学代写mechanics代考|Experimental Arrangement

The experimental arrangement of the transmission CGS method is shown in Fig. 5.6a. The specimen is illuminated by a collimated beam of coherent light. The transmitted beam is then incident normally on two diffractions gratings of the same pitch and parallel rulings placed on planes parallel to the specimen. A filtering lens $L_1$ is placed after the two gratings. The frequency content of the lens is displayed on its focal plane. Either of the $\pm 1$ diffraction orders is blocked by a filtering aperture. The light passing through the filter is obtained on the image plane of the lens $L_2$. For opaque materials, the above optical arrangement is modified by using a beam splitter (Fig. 5.6b). The reflecting surface of the specimen is illuminated by a collimated beam of coherent light using a beam splitter. The reflected beam passes through the beam splitter, and as in the previous case, then passes through the two diffraction gratings, the lens, the filter plane, and is obtained on the image plane of a lens.

The equations of the image obtained in the above optical arrangement of the transmission or reflection CGS method will follow the developments of Sect. $4.2$ using a simplified two-dimensional analysis [9]. For simplicity, it is assumed that the diffraction gratings have a sinusoidal transmission so that three diffraction rays are obtained at each grating.

First, consider that the specimen is undeformed and assume that the filtering lens blocks all but the $-1$ diffraction order. Following the analysis of Sect. $4.2$, the diffracted light ray $r_{-1,0}(-1$ diffraction order from the first grating, zero diffraction order from the second grating) deviates by an angle $\theta$ given by $\sin \theta \approx \theta=\lambda / p$, where $p$ is the pitch of the grating and $\lambda$ is the wavelength of light (Eq. (2.50)) (Fig. 5.7a). The deflected ray $r_{-1,0}$ exits the second grating at zero order with no deviation. The ray $r_{-1,0}$ makes an angle $\theta$ with the normal to the gratings. Consider now the ray $r_{0,-1}$ that exits at first grating at zero order with no deviation angle and the second grating at $-1$ order with deviation angle $\theta$. Both angles are equal because the two gratings have the same pitch.

The rays $r_{-1,0}$ and $r_{0,-1}$ are parallel, and therefore, interfere. Let the magnitudes of the electric vectors of the two rays are given by (Eq. (2.25)).

## 物理代写|力学代写mechanics代考|General Equations for Reflecting Surfaces

Consider a reflecting surface referred to the system $O x y z$ with equation
$$z=f(x, y)$$
illuminated by a collimated light beam perpendicular to the plane $O x y$ (Fig. 6.1). When a reference screen is placed parallel to the plane $O x y$ at distance $z_0$ the deviation vector $\boldsymbol{w}$ of the reflected ray from a point $P(x, y)$ on the surface to a point $P^{\prime}(x, y)$ on the screen, according to Snell’s law of reflection, is given by
$$w=w_x \boldsymbol{i}+w_y \boldsymbol{j}$$

with
$$\begin{gathered} w_x=\left(z-z_0\right) \tan 2 \alpha, \quad w_y=\left(z-z_0\right) \tan 2 \beta \ \tan \alpha=\frac{\partial f(x, y)}{\partial x}, \quad \tan \beta=\frac{\partial f(x, y)}{\partial y} \ \tan 2 \alpha=\frac{2 \partial f(x, y) / \partial x}{1-\left[\frac{\partial f(x, y)}{\partial x}\right]^2}, \quad \tan 2 \beta=\frac{2 \partial f(x, y) / \partial y}{1-\left[\frac{\partial f(x, y)}{\partial y}\right]^2} \end{gathered}$$
where $i$ and $j$ are the unit vectors referred to the projection $O^{\prime} x^{\prime} y^{\prime}$ of the frame $O x y$ on the screen.

We refer vector $w$ to the origin of the system $O^{\prime} x^{\prime} y^{\prime}$. Then, the vector $W$ which defines the position of the image point $P^{\prime}\left(W_x, W_y\right)$ of point $P(x, y)$ of the surface on the plane $O^{\prime} x^{\prime} y^{\prime}$ of the screen is given by
$$\boldsymbol{W}=W_x \boldsymbol{i}+W_y \boldsymbol{j}=\boldsymbol{r}+\boldsymbol{w}$$
with
$$W_x=x+\left[f(x, y)-z_0\right] \frac{2 \partial f(x, y) / \partial x}{1-\left[\frac{\partial f(x, y)}{\partial x}\right]^2}$$ $W_y=y+\left[f(x, y)-z_0\right] \frac{2 \partial f(x, y) / \partial y}{1-\left[\frac{\partial f(x, y)}{\partial y}\right]^2}$

## 物理代写|力学代写mechanics代考|General Equations for Reflecting Surfaces

$$z=f(x, y)$$

$$w=w_x \boldsymbol{i}+w_y \boldsymbol{j}$$

$$w_x=\left(z-z_0\right) \tan 2 \alpha, \quad w_y=\left(z-z_0\right) \tan 2 \beta \tan \alpha=\frac{\partial f(x, y)}{\partial x}, \quad \tan \beta=\frac{\partial f(x, y)}{\partial y} \tan 2 \alpha=$$

$$\boldsymbol{W}=W_x \boldsymbol{i}+W_y \boldsymbol{j}=\boldsymbol{r}+\boldsymbol{w}$$

$$W_x=x+\left[f(x, y)-z_0\right] \frac{2 \partial f(x, y) / \partial x}{1-\left[\frac{\partial f(x, y)}{\partial x}\right]^2}$$
$$W_y=y+\left[f(x, y)-z_0\right] \frac{2 \partial f(x, y) / \partial y}{1-\left[\frac{\partial f(x, y)}{\partial_y}\right]^2}$$

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