解决镜像和放大公式问题
双筒望远镜通常用于观察远处的物体。在通过这个设备观察时,我们注意到通过双筒望远镜看到的物体的大小与我们用肉眼观察到的不同。这是因为双筒望远镜将来自远处物体的光线集中起来,使我们能够通过使它们看起来更大来更详细地看到它们。这是镜子公式和放大倍数的演示。
镜像公式
镜面公式显示了球面镜的物距、像距和焦距之间的关系。因此,公式如下,
1/u + 1/v = 1/f
在哪里,
- u 是物距,
- v 是图像距离,并且
- f是镜子的焦距
While solving numerical problems using mirror formula one must remember two important things according to the sign convention for spherical mirrors, that are:
- If object is at left side of the mirror the object distance is taken as negative, if it is in right it is positive.
- For focal length the sign is depend on type of mirror we are using if mirror is concave then focal length is taken negative and if mirror is convex then focal length is positive.
根据曲率半径的镜像公式:
因为曲率半径 (R) 是焦距 (f) 的两倍,即
R = 2f
要么
f = R/2
因此,镜像公式可以写成:
1/u + 1/v = 1/f = 2/R
放大
大家一定注意到了,凹面镜或凸面镜只是简单地放大或缩小图像的大小,这是对物体的放大。
- 放大率定义为图像高度 (h i ) 与物体高度 (h o ) 的比值。
放大倍率 = 图像高度 / 物体高度
要么
m = H i / H o
放大倍率也可以定义为像距 (v) 与物距镜子 (v) 的比值。
放大倍率 = -像距/物距
要么
m = -v / u
Determination of the Nature of the Image formed by a spherical mirror:
- The positive magnitude of magnification shows us that a virtual and erect image is formed.
- The negative magnitude of magnification shows us that a real and inverted image is formed.
镜像公式和放大公式的问题
问题1:将物体放置在距凸面镜极点2倍焦距处,计算线性放大率。
解决方案:
Let the Focal length of mirror = f
So, the object distance, u = -2f
The formula to calculate image distance we use mirror formula as,
1 / v + 1 / u = 1 / f
Therefore,
1 / v + 1 / -2f = 1 / f
1 / v = 1 / f + 1 / 2f
= 3 / 2f
or
v = 2f / 3
Magnification is given as,
m = – v / u
= -(2f/3) / (-2f)
= 1/3
问题 2. 如果图像距离为 6 厘米,物体在凹面镜前 12 厘米处,计算形成的放大倍率。
解决方案:
Given that,
The distance of object, u = – 12 cm
The distance of image, v = – 6 cm
Since,
Magnification is given by,
m = – v / u
Therefore,
m = – (-6 / -12)
= -0.5
Hence, the image will be diminished by nearly half as size of object.
问题3:在实验中图像的高度是12厘米,而物体的高度是3厘米,你会确定形成的放大倍数吗?
解决方案:
Given that,
Height of image = 12 cm
Height of object = 3 cm
Magnification in terms of height is given by,
m = height of image / height of object
= 12 / 3
= 4
Therefore magnification is 4.
问题4:在凹面镜的情况下,如果将物体放置在12厘米的距离处。如果物像比为 1:2,则确定与镜子的图像距离。
解决方案:
Given that,
The object distance, u = -12 cm
Ratio of object to image height = 1/2
Magnification = height of image / height of object
= 1/ (1/2)
= 2
Now, magnification in terms of distance of object and image from the mirror,
m = – v / u
= – v / -12
2 = v / 12
or
v = 12 × 2
= 24
Therefore the distance of image from the mirror is equal to 24 cm.
问题5:如果物距为18cm,图像距凹面镜的距离为6cm,计算所形成图像大小的变化。
解决方案:
Given that,
The object distance, u = -12 cm
Image distance, v = – 6cm
Magnification, m = – v/u
= – (-6 / -18)
= -1/3
which means that size of image is 1/3rd of the size of object.
问题6:货车后视凸面镜的曲率半径为6m。如果汽车距离卡车后视镜 8 m。计算形成图像的距离。
解决方案:
Given that,
Radius of curvature, R = 6 m
Object distance, u = -8 m
Focal length is half of Radius of curvature,
f = R/2
= 6/2
= 3 m
Using mirror formula
1 / v + 1 / u = 1 / f
1 / v + 1 / -8 = 1 / 3
1 / v = 1 / 3 + 1 / 8
= 11 / 24
v = 24 / 11 m
The image is formed at distance of 24 / 11 behind the mirror.
问题 7:凹面镜产生的图像大小为物体的 n 倍,焦距为 f。如果图像是真实的,则找到物体到镜子的距离。
解决方案:
Given that
Size of image = n × size of object
n = Size of image / size of object = magnification
Since the image is real, it must be inverted hence magnification will be negative,
m = -n
Let d is the distance of object then,
m = -v/u
-n = -v / d
or
v = nd
Therefore, the mirror formula:
1 / f = 1/v + 1/u
becomes,
1/f = 1/nd + 1/d
or
1/f = 1/d(1/n + 1)
or
1/d = n/ f(n + 1)
Therefore,
d = f (n + 1)/ n
问题 8:物体应该放在哪里才能获得 1/3 的放大倍率?如果将物体放置在距凸面镜 60 厘米处,则产生的放大倍数为 1/2。
解决方案:
Given that,
u = -60 cm
m = 1/2
So,
-v/u = 1/2
and
v/60 = 1/2
or
v = 30 cm
Since, the mirror formula is:
1 / v + 1 / u = 1 / f
Therefore,
1 / 30 + 1 / (-60) = 1/f
1/f = ( 2-1 ) / 60 = 1 / 60
f = 60 cm
Now for magnification = 1 / 3,
– v / u = 1 / 3
or
v = – u / 3
using mirror formula
1 / v + 1 / u = 1 / f
1 / (-u/3) + 1/ u = 1/ 60
-3/ u + 1/u = 1/60
-2/ u = 1/60
or
u = -120 cm
object should be placed at 120 cm in front of mirror to get magnification of 1/3.
问题9:在凹面镜的情况下,如果物距是11cm,那么它的焦距是11cm,计算像距。
解决方案:
Given that,
Distance of object, u = -11 cm
Focal length, f = -11cm
Using mirror formula,
1 / v + 1 / u = 1 / f
Therefore,
1 / v + 1 / -11 = 1/ -11
So,
1/v = 0
or
v = infinity
This means that image will be at infinity if object is present at the focal length.
问题10:如果物距在凹面镜前32厘米,则镜面焦距为16厘米。说明所形成图像的性质和大小。
解决方案:
Given that,
Object distance, u = -32 cm
Focal length , f = -16 cm
For image distance use mirror formula,
1 / v + 1 / u = 1 / f
Therefore,
1/ v + 1/ -32 = 1/ -16
or
1/ v = 1/ -16 + 1/ 32
or
1/ v = -1 / 16
So,
v = -16 cm
Hence the image is located 16 cm in front of the mirror. and the image formed is real and inverted.
Size of image will be same as that of object, as it is located at center of curvature.