什么是镜头公式和放大倍率？

光是一种可以用肉眼看到的能量。我们主要通过使用光来观察物体并了解我们周围的世界。光以大约 3 × 10 ⁸ ms 的极快速度沿直线传播。小光源会在不透明物体上产生强烈的阴影。这意味着光线沿直线传播，路径称为一束光线，一组光线称为一束光线。

光线指示光传播的方向。当光线照射到两个透明介质之间的表面时，它会反射和折射，导致光线弯曲。光线也在障碍物的边缘弯曲，尽管由于光辐射的波长非常短，弯曲相对较小。这被称为光衍射。

镜头制造商方程

The image distance can be computed using the lens formula and knowledge of the object distance and focal length. The Lens formula describes the relationship between the distance of an image I the distance of an object (o), and the focal length (f) of the lens in optics. The lens formula works for both convex and concave lenses.

$\begin{aligned}\dfrac{1}{v}-\dfrac{1}{u}&=\left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\end{aligned}$

镜头制造商使用镜头方程来制造具有所需焦距的镜头。具有不同焦距的镜头用于各种光学设备。透镜的焦距由两个表面的曲率半径和透镜材料的折射率决定。

考虑一个折射率为 μ ₂的薄玻璃透镜，曲率中心为 C ₁和 C ₂ ，曲率半径为 R ₁和 R ₂ 。令 μ ₁表示周围介质的折射率。当点物体'O'保持在距镜头距离u的轴上时，光线OP无偏差地通过光学中心。如果另一条射线 OA 没有与 AB 一起被第一个表面折射，它将到达点 I'。如果第二个表面不存在。但是，由于第二个表面，光线在 B 点再次发生折射并到达 I 点。

通过凸透镜折射

For the first refraction, the object distance is u, and the image distance is v’. By using the relation among u. v. μ₁, and R for refraction at a single curved (convex) surface.

$\dfrac{\mu_2}{v'}-\dfrac{\mu_1}{u}=\dfrac{(\mu_2-\mu_1)}{R_1}$                                                                                                                               ……(1)

For the second refraction, I’ acts as a virtual object in lens medium of refractive index μ₂, to produce a final image at I in surrounding medium of refractive index μ_1. By using the relation among u. v. μ₁, and R for refraction at a single curved (convex) surface.

$\dfrac{\mu_1}{v} - \dfrac{\mu_2}{v'} = \dfrac{(\mu_1 - \mu_2)}{R_2}$                                                                                                                         ……(2)

Adding the Equation (1) and (2), we get

$\begin{aligned}\dfrac{\mu_1}{v} - \dfrac{\mu_2}{u}&=\dfrac{\mu_2 - \mu_1}{R_1} + \dfrac{\mu_1 - \mu_2}{R_2}\\\dfrac{1}{v}-\dfrac{1}{u}&=\left(\dfrac{\mu_2 - \mu_1}{\mu_1}\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\\&=\left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\end{aligned}$                                                                                                                                                                              ……(3)

• Case 1: If the point object is at infinity, its image is formed at the focal point. i.e. if u = ∞, then v = f; therefore equation (3) becomes-

$\dfrac{1}{f} = \left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$                                                                                   ……(4)

This is the lens maker`s equation. The lens of any desired focal length can be produced by choosing proper values of R₁, R₂, μ₁, μ_2.

• Case 2: When the lens is kept in an air medium, μ₁  = 1 and the equation (4) becomes –

$\dfrac{1}{f} = ({\mu_2} - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$                                                                                                                       ……(5)

Using the new Cartesian sign convention, we get a positive focal length for the convex lens and a negative focal length for the concave lens.

Comparing the equation (3) and (5), we get

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

This is the lens formula,  which is used to find the focal length of a given lens practically.

放大

放大意味着使物体看起来比实际更大。以下是确定不同情况的放大倍数的不同情况：

镜头产生的放大倍率（m）

镜头的放大倍率 (m) 定义为图像高度与物体高度的比值。

m = 图像高度 / 物体高度

m = h _i / h _o

它也以像距和物距的形式呈现。

m = 图像距离/物距

m = v / u

它相当于图像与物体的距离比。

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凸透镜产生的放大倍率

我们知道凸透镜的特性是虚的和正的。因为凸透镜可以创建虚拟和实际图像，所以凸透镜产生的放大倍率可以是正的也可以是负的。放大对虚拟图像有利，但对真实图像不利。即虚拟图像的正（+ve）和真实图像的负（-ve）。

情况1：如果放大倍数小于1，则表示图像小于物体。 |m|<1 ，图像被缩小。

情况2：如果放大倍数大于1，则图像大于物体。

|m|>1 ，图像被放大。

情况3：如果放大倍数为1，则图像与物体大小相同。 |米| = 1 ，图像与对象大小相同。

通过凹透镜放大

当物体在凹透镜前面时，图像在同一侧的同一物体前面。凹透镜总是产生一个虚拟的、正立的和缩小的图像。因为凹透镜总是产生虚像，所以它们实现的放大倍率总是正的，它总是产生比物体小的图像。

放大倍数小于一，所以图像比物体小。

|m|<1 ，图像被缩小。

下面的表格表示表示不同镜头的不同情况下图像的放大率和性质，如下所示：

Lens	Properties	Object distance (u)	Image distance (v)	Magnitude in terms of ‘u’ and ‘v’	Magnitude in terms of image height and object height
Convex Lens	Image produced are virtual and  upright(real image)	-ve	Virtual Image: -ve Real Image: +ve	Virtual Image: +ve Real Image: -ve	|m|=1 |m|<1 |m|>1	when hi=ho when hio  when hi>ho
Concave	Image produced is always virtual	-ve	-ve	Virtual image: +ve	|m|<1	–

示例问题

问题一：薄镜片浸入水中，焦距会发生变化吗？

回答：

The focal length of a lens is determined by the refractive index of the lens’s material in relation to its surroundings, as determined by the lens maker’s formula.

$\frac{1}{f} = (\frac{\mu_2}{\mu_1} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

 When a thin lens is submerged in water, its relative refractive index decreases, and therefore its focal length increases (and the lens’s power decreases).

Let f_a be the focal length of a thin glass lens in air and f_w be the focal lengths of a thin glass lens in water.

 Let _aμ_g and _wμ_g = _aμ_g / _aμ_w  be the relative refractive indices of glass with respect to air and water.

 Then,

$\frac{1}{f_a} = (_a\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

$\frac{1}{f_w} = (_w\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

Dividing both equations we get,

$\frac{f_w}{f_a} = \frac{_a\mu_g - 1}{_w\mu_g - 1}$

Since, _aμ_g > _wμ_g , f_w > f_a 

$f_w = \frac{_a\mu_g - 1}{_w\mu_g - 1} \times f_a$

问题2：定义透镜的焦距，给出凹凸透镜焦距的符号？

回答：

The focal length of a lens is the distance between its optical center and its primary focus.

It is defined as the distance between the optical center and the second primary focus, so that the focal length of a convex lens is positive and that of a concave lens is negative. 

问题3：镜头产生的线性放大率是多少？它的标志有什么意义？

回答：

• The linear magnification produced by a lens is the factor by which the image size is changed from the object size.

           It is defined as the ratio of the height of an image to the height of an object.

                                      m = Height of the image / Height of the object

m = hi / ho

• The linear magnification indicator specifies whether the image is upright or inverted in relation to the object. The magnification of an upright picture is positive, whereas the magnification of an inverted image is negative.

示例问题

问题1：凸透镜的曲率半径为40cm和50cm，如果其材料的折射率为2.1，计算焦距。

解决方案：

Given that: Refractive index (μ) = 2.1

Radii R₁ = 40cm

Radii R₂ = -50cm    (sign conventions)

$\frac{1}{f} = (\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$   

                = (2.1 – 1) (1/40 – 1/(-50))

                = 1.1 × 9 / 200

                = 0.0495

       1 / f  = 1 / 0.0495

               = 20.2020 cm

The focal length of convex lens is 20.2020 cm.

问题 2：凸透镜在距透镜 40 厘米处形成物体的实像和倒像。如果图像与物体大小相同，那么在凸透镜前放置的物体会在哪里？

解决方案：

Given that: Image distance (v) = 40cm

Lens type is convex and image is real image

Height of image is as same size as of object i.e., h_i = -h_o

Since, magnification (m) = h_i / h_o = v / u

                                       = -h_o / h_o = v / u

                             v / u = -1    

                                  u = -1 × v

                                u  =  -40cm.

1/f = 1/v -1/u

      = 1/40 – 1/(-40)

      = 2/40

1/f = 1/20

f = 20cm = 0.2m

The object is placed 40 cm away from the convex lens.

问题 3：焦距为 20cm 的凹透镜在距透镜 15cm 处形成针状图像。针距镜头多远？

Solution:

Given that:  

Focal length (f) = -20cm

Image distance (v) = -15 cm

1/f = 1/v – 1/u

or

1/u = 1/v – 1/f

      = 1/(-15) – (1/-20)

      = -20 + 15 / 300

      = – (1/60)cm

   u = -60 cm

The object is placed 60cm in front of the concave lens.

问题4：由玻璃制成的凹透镜在空气中的焦距为20cm。浸入水中时找到它的焦距。假设玻璃镜片的折射率为 1.5，水中的折射率为 4。

解决方案：

Let f_a be the focal length of lens in air

Let f_w be the focal length of lens in water.

Given that:  

f_a = – 20cm (concave lens)

_aμ_g = 1.5

_aμ_w = 4

_wμ_g = _aμ_g / _aμ_w

      = (3/2) / (4/1) 

      = 3/8

$\frac{1}{f_a} = (_a\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

$\frac{1}{f_w} = (_w\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

 Dividing equation both the equations we get,

 $\frac{f_w}{f_a} = \frac{_a\mu_g - 1}{_w\mu_g - 1}$      

 $f_w = \frac{_a\mu_g - 1}{_w\mu_g - 1} \times f_a$ 

       = (1.5 – 1) / (0.375 – 1) × (-20)

       = 0.5 / (-0.625) × (-20)

f_w = 16cm

问题 5：光学系统使用两个薄凸透镜相接触，有效焦距为 30/4 cm。如果其中一个镜头的焦距为 30 厘米，请找出另一个镜头的焦距。

解决方案：

Given that:

f = 30/4cm

f₁ = 30cm    

1/f = 1/f₁ + 1/f₂

1/f₂ = 1/f – 1/f₁

       = 1/(30/4) – 1/30

       = 4/30 – 1/30

1/f₂ = 1/10

The focal length of the other lens, f₂ is equal to 10 cm.