91影视

1. The index of refraction $n$of a medium is defined by the ratio

$n=\frac{c}{谓}$

where c =data-custom-editor="chemistry" $3.0脳{10}^8$m/s is the speed of light in vacuum and $谓$is the speed of light in the medium.

2.The Ray Model of Light:

Light travels along straight lines, called light rays, at speed v=c/n, where n is the material's index of refraction. A light ray continues forever unless an interaction with matter causes it to reflect, refract, scatter, or be absorbed. Light rays come from objects. Each point on the object sends rays in all directions. The eye sees an object (or an image) when diverging rays are collected by the pupil and focused on the retina. Ray optics is valid when lenses, mirrors, and apertures are larger than 鈮� 1 mm.

3.The average speed of a particle is equal to the ratio of the total distance it travels to the total interval during which it travels that distance:

${谓}_{avg}=\frac{d}{鈭�t}$

• The light ray that travels from point A to point B.

• The ray crosses the boundary at position $x$

• The angle of incidence of the light ray is:${胃}_1$

• The angle of refraction of the light ray is:${胃}_2$.

• The index of refraction of the medium for which the light ray is incident is: $n_1$

• The index of refraction of the medium for which the light ray is refracted is:$n_2$.

• We are not allowed to use Snell's law.

• $x_{min}$is the value of xx for which the light travels from A to B in the shortest possible time.

find an expression for the time tt taken by the light ray to travel from point A to point B.

The speed $v_1$of the light ray in medium 1 is found in terms of its index of refraction $n_1$from Equation (1):

$v_1=\frac{c}{n_1}$

Similarly, the speed $v_2$of the light ray in medium 2 is:

$v_2=\frac{c}{n_2}$

The time of travel of the light ray from point A to the boundary is found from Equation (2):

$$\begin{gathered} t_1=\frac{d_1}{v_1} \\ =\frac{d_1}{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.}} \\ =\frac{n_1{d}_1}{c} \end{gathered}$$

where $d_1$is the distance traveled by the light ray from point A to the boundary. $d_1$is found in terms of aa and x by applying the

Pythagorean theorem to the yellow-shaded triangle in Figure 1:

$d_1=\sqrt{x^2+b^2}$

Substitute for $d_1$into Equation (3):

role="math" localid="1650291131743" $t_1=\frac{n_1\sqrt{x^2+a^2}}{c}$

The time of travel of the light ray from the boundary to point B is found from Equation (2):

_{$$\begin{gathered} t_2=\frac{d_2}{v_2} \\ =\frac{d_2}{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.}} \\ =\frac{n_2{d}_2}{c} \end{gathered}$$}

where $d_2$is the distance traveled by the light ray from the boundary to point B. $d_2$is found in terms of x, w, and b by applying the Pythagorean theorem to the green-shaded triangle in Figure 1:

$d_2=\sqrt{{(w-x)}^2+b^2}$

Substitute for $d_2$into Equation (4):

$t_2=\frac{n_2\sqrt{{(w-x)}^2+b^2}}{c}$

The time $t$taken by the light ray to travel from point A to point B is then:

t=role="math" localid="1650292064095" $$\begin{gathered} t=t_1+t_2 \\ =\frac{n_1\sqrt{x^2+a^2}}{c}+\frac{n_2\sqrt{{(w-x)}^2+b^2}}{c} \\ =\frac{1}{c}\left[n_1\sqrt{x^2+a^2}+n_2\sqrt{{(w-x)}^2+b^2}\right] \end{gathered}$$

determine an expression from which$x_{min}$ can be found.

The expression from which $x_{min}$can be determined is found by differentiating $t$with respect to x and then setting the derivative equal to zero. So,

$$\begin{gathered} \frac{dt}{dx}=0 \\ \frac{d}{dx}\left\{\frac{1}{c}\left[n_1\sqrt{x^2+a^2}+n_2\sqrt{{(w-x)}^2+b^2}\right]\right\}=0 \\ \frac{1}{c}\frac{d}{dx}\left[n_1\sqrt{x^2+a^2}+n_2\sqrt{{(w-x)}^2+b^2}\right]=0 \\ \frac{d}{dx}\left\{n_1{\left(x^2+a^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+n_2{\left[{(w-x)}^2+b^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}=0 \\ \frac{1}{2}{n}_1{\left(x^2+a^2\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2x\right)+\frac{1}{2}{n}_2{\left[{\left(w-x\right)}^2+b^2\right]}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2\right)\left(w-x\right)\left(-1\right)=0 \\ \frac{n_{1}x}{\sqrt{x^2+a^2}}-\frac{n_2\left(w-x\right)}{\sqrt{{\left(w-x\right)}^2+b^2}}=0 \\ \frac{n_{1}x}{\sqrt{x^2}+a^2}=\frac{n_2(w-x)}{\sqrt{{(w-x)}^2+b^2}} \end{gathered}$$

The value of $x$can determined using the above expression.

derive Snell's law from the geometry of the figure and the answer to part (b).

From the geometry of Figure 1, we see that the cosine of the angle ${90}^{慰}-{胃}_2$is:

role="math" localid="1650293138594" $$\begin{gathered} \cos ({90}^{慰}-{胃}_2)=\frac{x}{d_1} \\ \cos ({90}^{慰}-{胃}_2)=\frac{x}{\sqrt{x^2+a^2}} \end{gathered}$$

where $\cos ({90}^{慰}-{胃}_2)$is equal to the sine of ${胃}_2$'s complementary angle ${胃}_1\sin {胃}_1$. So

$\sin {胃}_1=\frac{x}{\sqrt{x^2+a^2}}$

The sine of the angle ${胃}_2$is also found from the geometry of the green-shaded triangle in Figure 1:

$$\begin{gathered} \sin {胃}_2=\frac{w-x}{d_2} \\ =\frac{w-x}{\sqrt{{(w-x)}^2+b^2}} \end{gathered}$$

Substituting for $\sin {胃}_{1}and\sin {胃}_2$from Equations (6) and (7) into Equation (5), we obtain Snell's law:

$n_1\sin {胃}_1=n_2\sin {胃}_2$