凝聚镜像

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Concave mirrors are used in a number of applications. They can form upright, enlarged images, and are therefore useful in makeup application or shaving. They are also used in flashlights and headlights because they can project parallel beams of light, and in telescopes because they can focus light allowing for clearer and magnified images of distant objects.
::凝聚镜像用于许多应用中,它们可以形成直立、放大的图像,因此在化妆或剃须方面有用。它们也可以用于手电筒和头灯,因为它们可以投射平行光束,也可以用于望远镜,因为它们可以聚焦光线,可以清晰放大远方物体的图像。

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The photograph above shows the grinding of the primary mirror in the Hubble space telescope. The Hubble Space Telescope is a reflecting telescope with a mirror approximately eight feet in diameter, and was deployed from the Space Shuttle Discovery on April 25, 1990.
::以上照片显示了哈勃空间望远镜主镜的研磨情况,哈勃空间望远镜是一个反射望远镜,直径约为8英尺,是1990年4月25日从航天飞机发现号上部署的。

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Image in a Concave Mirror
::图像,在收藏镜中

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Reflecting surfaces do not have to be flat. The most common curved are spherical. A spherical mirror is called a concave mirror if the center of the mirror is further from the viewer than the edges are. A spherical mirror is called a convex mirror if the center of the mirror is closer to the viewer than the edges are.
::反射面不必是平的。 最常见的曲线是球形的。 球面的镜面被称为共形镜, 如果镜面的中心比边缘更远的话。 如果镜面的中心比边缘更近, 球面的镜面就叫锥形镜。 如果镜面的中心比边缘更近的话, 球面的镜面就叫锥形镜 。

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To see how a concave mirror forms an image, consider an object that is very far from the mirror so that the incoming rays are essentially parallel. For an object that is infinitely far away, the incoming rays would be exactly parallel. Each ray would strike the mirror and reflect according to the law of reflection ( angle of reflection is equal to the angle of incidence). As long as the section of mirror is small compared to its radius of curvature, all the reflected rays will pass through a common point, called the focal point ( f) .
::要看到一个混凝土的镜子如何形成图像, 请考虑一个离镜子很远的物体, 这样进入的射线基本上是平行的。 对于一个无限遥远的物体, 进入的射线将是完全平行的。 每个射线会按照反射法则( 反射线的角等于事件角度) 照镜子并反射。 只要反射线的角小于其卷曲半径, 所有反射的射线都会通过一个共同点, 称为焦点( f ) 。

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If too large a piece of the mirror is used, the rays reflected from the top and bottom edges of the mirror will not pass through the focal point and the image will be blurry. This flaw is called spherical aberration and can be avoided either by using very small pieces of the spherical mirror or by using parabolic mirrors.
::如果反射镜使用过大, 反射镜上下边缘的射线不会通过焦点, 图像会模糊。 这个瑕疵被称为球体畸变, 可以通过使用极小的球面镜片或抛光镜来避免。

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A line drawn to the exact center of the mirror and perpendicular to the mirror at that point is called the principal axis . The along the principal axis from the mirror to the focal point is called the focal length . Since the focal point is on the same side of the mirror as the object, the focal length is assigned a positive number. The focal length is also exactly one-half of the radius of curvature of the spherical mirror. That is, if the spherical mirror has a radius of 8 cm, then the focal length will be 4 cm.
::绘制到镜中正中线的直角线与镜中正反射线的垂直线称为主轴。主轴从镜中到焦点的长度称为焦线长度。由于焦点与对象在镜子的同一侧面,焦线长度被指定为正数。焦线长度也是球面镜半径的半径。也就是说,如果球面镜半径为8厘米,则焦线长度为4厘米。

 

 

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Objects Outside the Center Point
::中心点外物体

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Above is a spherical mirror with the principal axis, the focal point, and the center of curvature ( C)  identified on the image. An object has been placed well beyond C,  and we will treat this object as if it were infinitely far away. There are two rays of light leaving any point on the object that can be traced without any drawing tools or measuring devices. The first is a ray that leaves the object and strikes the mirror parallel to the principal axis that will reflect through the focal point. The second is a ray that leaves the object and strikes the mirror by passing through the focal point; this ray will reflect parallel to the principal axis.
::上面是主轴、 焦点和曲轴中心( C) 的球面镜。 一个对象被放置在 C 以外的地方, 我们将把这个对象看成是远不远的。 有两线光线, 使对象上的任何点在没有任何绘图工具或测量装置的情况下可以追踪。 第一个是将对象留下的射线, 并击中镜与主轴平行的射线, 该主轴将通过焦点反射。 第二个是离开对象的射线, 并通过焦点撞击镜; 这个射线将与主轴平行反射 。

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These two rays can be seen in the image below. The two reflected rays intersect after at a point between C  and F . Since these two rays come from the tip of the object, they will form the tip of the image.
::这两个射线可以在下面的图像中看到。两个反射射线在C和F之间的一个点相交。由于这两个射线来自天体的顶端,它们将形成图像的顶端。

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If the image is actually drawn in at the intersection of the two rays, it will be smaller and inverted, as shown below. Rays from every point on the object could be drawn so that every point could be located to draw the image. The result would be the same as shown here. This is true for all concave mirrors with the object outside C : the image will be between C  and F , and the image will be inverted and diminished (smaller than the object). 
::如果图像实际上是在两个射线的交叉点上绘制的,则图像将变小和反转,如下文所示。对象上每个点的射线可以绘制,以便每个点的位置都能够绘制图像。结果将与此处显示的结果相同。对于C外对象的所有相形镜都是如此:图像将在 C 和 F 之间,图像将被反转和缩小(比对象小)。

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The heights of the object and the image are related to their distances from the mirror. In fact, the ratio of their heights is the same ratio as their distances from the mirror. If $\begin{align*}d_o\end{align*}$  is object distance, $\begin{align*}d_i\end{align*}$  the image distance, $\begin{align*}h_o\end{align*}$  the object height and $\begin{align*}h_i\end{align*}$  the image height, then
::对象的高度和图像与它们距离镜像的距离有关。事实上,它们的高度比与它们距离镜子的距离比例相同。如果是对象距离,那么,如果是图像距离,则按对象高度和图像高度的高度,然后

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$$\begin{align*}\frac{h_o}{h_i}=\frac{d_o}{d_i}.\end{align*}$$
::霍希多迪。 \hhi=dodi。

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It can also be shown that $\begin{align*}\frac{d_o}{d_i}=\frac{d_o-f}{f}\end{align*}$  and from this, we can derive the mirror equation ,
::也可以证明 dodi=do -ff, 从中我们可以得出镜面方程式,

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$$\begin{align*}\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}.\end{align*}$$
::1Do+1di=1f.

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In this equation, f  is the focal length d _o  is the object distance, and d _i  is the image distance.
::在此方程中, f 是焦距是天体距离, di 是图像距离 。

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The magnification equation for a mirror is the image size divided by the object size, where  m  gives the magnification of the image. 
::镜像的放大方程式是图像大小除以对象大小, m 给图像放大。

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$$\begin{align*}m=\frac{-d_i}{d_o}\end{align*}$$
::mdido

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Example 1  
::例1

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A 1.50 cm tall object is placed 60.0 cm from a concave mirror whose radius of curvature is 30.0 cm. What is the image distance and what is the image height?
::1.50厘米高的天体从圆形镜中放置60.0厘米,其半径为30.0厘米。 图像距离是多少,图像高度是多少?

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Recall the equation: $\begin{align*} \frac{1}{d_O} + \frac{1}{d_i} =\frac{1}{f} \end{align*}$
::回顾方程:1dO+1di=1f

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The focal length is one-half the center of curvature, so it is 15.0 cm. Substitute the known values into the equation:   $\begin{align*}\frac{1}{60.0 \text{ cm}} + \frac{1}{d_i} =\frac{1}{15 \text{ cm}}\end{align*}$  . Rearrange for   $\begin{align*}\frac{1}{d_i}\end{align*}$  : $\begin{align*}\frac{1}{d_i} =\frac{1}{15 \text{ cm}} - \frac{1}{60.0 \text{ cm}} = \frac{3}{60.0 \text{ cm}}\end{align*}$  . Take the reciprocal of both sides: $\begin{align*}d_i = \frac{60.0 \text{ cm}}{3} = 20cm\end{align*}$ .
::焦距是曲线中心的一半, 所以是15. 0厘米。 将已知值替换为方程 : 160.0 cm+1di=115 cm。 1di 的重新排列:1di = 115 cm-160 cm= 360.0 cm。 以两边的对等值 : di= 60.0 cm3= 20cm 。

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The image distance is 20.0 cm. The image is 3 times closer to the mirror than the object so it will be 3 times as small, or 0.5 cm tall.
::图像的距离是20.0厘米。图像离镜子更近3倍于对象3倍,因此是小3倍或0.5厘米高。

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Objects Between the Center Point and the Focal Point
::中心点与协调中心之间的物体

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Regardless of where the object is, its image's size and location can be determined using the equations given earlier in this section. Nonetheless, patterns emerge in these characteristics. We already know that the image of an object outside the center point is closer and smaller than the object. When an object is between the center point and the focal point, the image is larger and closer. These characteristics can also be determined by drawing the rays coming off of the object; this is called a  ray diagram. 
::无论对象在哪里,其图像的大小和位置都可以使用本节前面给出的方程式来确定。 尽管如此, 这些特性中会出现模式。 我们已经知道, 中心点以外的对象的图像比对象更近、更小。 当一个对象位于中心点和焦点点之间时, 图像会越大、 越近。 这些特性也可以通过从对象上绘制射线来确定; 这被称为射线图 。

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Look again at the image above that was shown earlier in the lesson. If you consider the smaller arrow to be the object and follow the rays backward, the ray diagram makes it clear that if an object is located between the center point and focal point, the image is inverted, larger, and at a greater distance.
::再看看上方的图像,上面的图像在课中稍早显示过。如果您认为小箭头是对象,并跟随射线向后,那么光线图显示,如果一个对象位于中心点和焦点点之间,则图像是反向的,更大,距离更大。

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Example 2  
::例2

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If an 3.00 cm tall object is held 15.00 cm away from a concave mirror with a radius of 20.00 cm, describe its image.
::如果一个3.00厘米高的物体在距离半径为2 000厘米的圆形镜像15.00厘米处,请描述其图像。

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To solve this problem, we must determine the height of the image and the distance from the mirror to the image. To find the distance, use the mirror equation:
::要解决这个问题,我们必须确定图像的高度以及从镜子到图像的距离。 要找到距离, 请使用镜像方程式 :

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$\begin{align*}\frac{1}{d_O} + \frac{1}{d_i} =\frac{1}{f}\end{align*}$  
::1dO+1di=1f

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$\begin{align*}\frac{1}{15.00 \text{ cm}} + \frac{1}{d_i} =\frac{1}{10.00 \text{ cm}}\end{align*}$  
::115.00厘米+1di=110.00厘米

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$\begin{align*}\frac{1}{d_i} =\frac{1}{10.00 \text{ cm}} - \frac{1}{15.00 \text{ cm}}\end{align*}$  
::1di=110.00厘米-15.00厘米

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$\begin{align*}d_i = 30 \text{ cm}\end{align*}$
::di=30厘米

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Next determine the height of the image:
::下一步决定图像的高度 :

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h o h i = d o d i = 3 h i = 15 30

$$\begin{align*}\frac{h_o}{h_i}=\frac{d_o}{d_i}=\frac{3}{h_i}=\frac{15}{30}\end{align*}$$
::霍希=多迪=3hi=1530

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Using this equation, we can determine that the height of the image is 6 cm.
::使用这个方程,我们可以确定图像的高度是6厘米。

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This image is a real image , which means that the rays of light are real rays. These are represented in the ray diagram as solid lines, while virtual rays are dotted lines.
::这个图像是真实的图像, 这意味着光线是真实的光线。 这些在光线图中作为实线表示, 而虚拟光线则是虚线 。

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Objects Inside the Focal Point
::协调中心内部的物体

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Consider what happens when the object for a concave mirror is placed between the focal point and the mirror. This situation is sketched at below.
::试想当凝聚镜的物体被放置在焦点和镜子之间时会发生什么。下面勾画了这种情况。

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Once again, we can trace two rays to locate the image. A ray that originates at the focal point and passes through the tip of the object will reflect parallel to the principal axis. The second ray we trace is the ray that leaves the tip of the object and strikes the mirror parallel to the principal axis.
::我们可以再次追踪两条射线以定位图像。 一条射线源自焦点并穿过天体顶端的射线将反射到主轴平行。 我们追踪的第二条射线是离开天体顶端并击倒镜与主轴平行的射线。

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Below is the ray diagram for this situation. The rays are reflected from the mirror and as they leave the mirror, they diverge. These two rays will never come back together and so a real image is not possible. When an observer looks into the mirror, however, the eye will trace the rays backward as if they had followed a straight line. The dotted lines in the sketch show the lines the rays would have followed behind the mirror. The eye will see an image behind the mirror just as if the rays of light had originated behind the mirror. The image seen will be enlarged and upright. Since the light does not actually pass through this image position, the image is virtual.
::光线从镜子中反射出来, 当它们离开镜子时, 它们会分裂。 这两个光线永远不会回到一起, 所以真实的图像是不可能的。 但是, 当观察者看到镜子时, 眼睛会像跟踪直线一样追踪光线向后。 草图中的虚线显示光线在镜子后面的线条。 眼睛会看到镜子后面的图像, 就像光线是从镜子后面开始的一样。 所看到的图像将会被放大和直立。 因为光线并没有真正穿过这个图像位置, 图像是虚拟的 。

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Example 3  
::例3

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A 1.00 cm tall object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm. Determine the image distance and image size.  
::1.00厘米高的天体从一个曲线半径为30.0厘米的剖面镜中放置10.0厘米。 确定图像的距离和图像大小。

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Since the radius of curvature is 30.0 cm, the focal length is 15.0 cm. The object distance of 10.0 cm tells us that the object is between the focal point and the mirror.
::由于曲线半径为30.0厘米,焦距为15.0厘米。10.0厘米的天体距离告诉我们,天体在焦距和镜子之间。

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$\begin{align*}\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}\end{align*}$  and plugging known values yields  $\begin{align*}\frac{1}{10.0}+\frac{1}{x}=\frac{1}{15}\end{align*}$
::1do+1di=1f 且插入已知值为 11.0+1x=115

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Multiplying both sides of the equation by $\begin{align*}30x\end{align*}$  yields $\begin{align*}3x + 30 = 2x\end{align*}$  and $\begin{align*}x = -30.0 \ cm\end{align*}$ .
::方程两侧乘以 30x 3x+30=2x 和 x30.0 cm 。

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The negative image distance indicates that the image is behind the mirror. We know the image is virtual because it is behind the mirror. Since the image is 3 times as far from the mirror as the object, it will be 3 times as tall. Therefore, the image height is 3.00 cm.
::负图像距离表示图像在镜子后面。 我们知道图像是虚拟的, 因为它在镜子后面。 由于图像距离镜的距离是对象的3倍, 图像的高度是对象的3倍。 因此, 图像的高度是 3 厘米 。

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A Cassegrain telescope uses a concave primary mirror to magnify light from very distant objects to form an image we can see. Play around with a Cassegrain telescope in the simulation below and see if you can until you get a clear view of Jupiter in the eyepiece:
::Cassegrain 望远镜使用一个圆锥镜主镜放大远方物体的光线以形成一个我们可以看到的图像。 在下面的模拟中, 与一个 Cassegrain 望远镜一起玩耍, 看看能否在眼睛上清晰地看到木星 :

 

 

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Summary
::摘要

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• A spherical mirror is concave if the center of the mirror is further from the viewer than the edges.
  ::球面镜是圆形的,如果镜子的中心离观察者更远,而不是边缘。
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• For an object that is infinitely far away, incoming rays would be exactly parallel.
  ::对于遥远的物体来说 进入的射线会完全平行
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• As long as the section of mirror is small compared to its radius of curvature, all the reflected rays will pass through a common point, called the focal point (F).
  ::只要镜片与曲线半径相比小,所有反射射线都会经过一个共同点,称为联络点(F)。
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• The distance along the principal axis from the mirror to the focal point is called the focal length, f. This is also exactly one-half of the radius of curvature.
  ::主轴从镜子到协调中心的距离称为焦距,f。这也正好是曲线半径的一半。
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• The mirror equation is $\begin{align*}\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}\end{align*}$ .
  ::镜像方程式为 1do+1di=1f。
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• The magnification equation is $\begin{align*}m=-\frac{d_i}{d_o}\end{align*}$ .
  ::放大方程式是mdido。
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• For concave mirrors, when the object is outside the center point ( C) , the image will be between C  and F  and the image will be inverted and diminished (smaller than the object).
  ::对于凝固镜像,当对象在中点(C)之外时,图像将在C和F之间,图像将被反转和缩小(小于对象)。
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• For concave mirrors, when the object is between C  and F , the image will be beyond C  and will be enlarged and inverted.
  ::对于相形镜来说,当对象在C和F之间时,图像将超过C,并将扩大和反向。
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• For concave mirrors, when the object is between F  and the mirror, the image will be behind the mirror and will be enlarged and upright.
  ::对于凝固的镜子来说,当物体在F和镜子之间时, 图像将在镜子后面, 并且会放大和直立。

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Review
::回顾

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1. A concave mirror is designed so that a person 20.0 cm in front of the mirror sees an upright image magnified by a factor of two. What is the radius of curvature of this mirror?
   ::雕刻镜面的设计是让一个在镜子前的20厘米的人看到一个直立的图像,其放大系数为2。这面镜子的曲缩半径是多少?
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2. If you have a concave mirror whose focal length is 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object, where should you place the object?
   ::如果您有一个焦距为 100.0 厘米的卷心镜子, 并且您想要一个直立和10.0倍高于对象的图像, 您应该把对象放在哪里 ?
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3. A concave mirror has a radius of curvature of 20.0 cm. Locate the image for an object distance of 40.0 cm. Indicate whether the image is real or virtual, enlarged or diminished, and upright or inverted.
   ::二次镜像的半径为20厘米。 定位对象距离为40厘米的图像。 显示图像是真实的还是虚拟的、 放大或缩小的、 直立或反转的 。
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4. A dentist uses a concave mirror to examine a tooth that is 1.00 cm in front of the mirror. The image of the tooth forms 10.0 cm behind the mirror.
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   1. What is the mirror’s radius of curvature?
      ::镜子的曲线半径是多少?
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   2. What is the magnification of the image?
      ::图像的放大作用是什么?
   
   ::牙医用卷轴镜检查镜子前1 000厘米的牙齿。 镜子后10. 0厘米的牙齿图象。 镜子的曲线半径是多少? 图像的放大度是多少?
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5. When a man stands 1.52 m in front of a shaving mirror, the image produced is inverted and has an image distance of 18.0 cm. How close to the mirror must the man place his face if he wants an upright image with a magnification of 2.0?
   ::当一个人站在1.52米长的剃须镜前时,所产生的图像是反向的,其图像距离为18厘米。 如果一个人想要一个直立的、放大2.0的图像,那么他必须贴近镜子多远?

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Explore More
::探索更多

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Use this resource to answer the questions that follow.
::使用此资源回答下面的问题 。

 

 

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1. What are the two main rays you need to draw a real image?
   ::要画出真实的图像, 您需要的两大光线是什么 ?
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2. What is the third ray? What can you use it for?
   ::第三射线是什么 你能用它做什么
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3. How do light rays through the focal point and vertex behave the same? How are they different?
   ::光线通过中心点和顶端如何表现相同?它们有什么不同?