An object O O^', 9.0 cm high, is 27 cm in front of a thin...

An object $O O^{\prime}, 9.0 \mathrm{~cm}$ high, is $27 \mathrm{~cm}$ in front of a thin concave lens of focal length $-18 \mathrm{~cm}$. Determine the position and height of its image $I I^{\prime}(a)$ by construction and $(b)$ by computation. (a) Choose the two convenient rays from $O$ shown in Fig. $38-6$. (1) A ray $O P$, parallel to the optical axis, is refracted outward in the direction $D$ as if it came from the principal focus $F$. (2) A ray through the optical center of the lens is drawn as a straight line $O C$. Then $I I^{\prime}$ is the image of $O O^{\prime}$. Images formed by concave or divergent lenses are virtual, erect, and smaller. (b) $\frac{1}{s_{o}}+\frac{1}{s_{i}}=\frac{1}{f}$ or $\frac{1}{27 \mathrm{~cm}}+\frac{1}{s_{i}}=-\frac{1}{18 \mathrm{~cm}} \quad$ or $\quad s_{i}=-10.8 \mathrm{~cm}=-11 \mathrm{~cm}$ Since $s_{i}$ is negative, the image is virtual, and it is $11 \mathrm{~cm}$ in front of the lens. $M_{T}=\frac{y_{i}}{y_{o}}=-\frac{s_{i}}{s_{o}}=-\frac{-10.8 \mathrm{~cm}}{27 \mathrm{~cm}}=0.40 \quad$ and so $\quad y_{i}=y_{o} M_{T}=(0.40)(9.0 \mathrm{~cm})=3.6 \mathrm{~cm}$ When $M_{T}>0$, the image is upright, and the same conclusion follows from the fact that $y_{i}>0 .$

An object $O O^{\prime}, 9.0 \mathrm{~cm}$ high, is $27 \mathrm{~cm}$ in front of a thin concave lens of focal length $-18 \mathrm{~cm}$. Determine the position and height of its image $I I^{\prime}(a)$ by construction and $(b)$ by computation.
(a) Choose the two convenient rays from $O$ shown in Fig. $38-6$.
(1) A ray $O P$, parallel to the optical axis, is refracted outward in the direction $D$ as if it came from the principal focus $F$.
(2) A ray through the optical center of the lens is drawn as a straight line $O C$.
Then $I I^{\prime}$ is the image of $O O^{\prime}$. Images formed by concave or divergent lenses are virtual, erect, and smaller.
(b) $\frac{1}{s_{o}}+\frac{1}{s_{i}}=\frac{1}{f}$ or
$\frac{1}{27 \mathrm{~cm}}+\frac{1}{s_{i}}=-\frac{1}{18 \mathrm{~cm}} \quad$ or $\quad s_{i}=-10.8 \mathrm{~cm}=-11 \mathrm{~cm}$
Since $s_{i}$ is negative, the image is virtual, and it is $11 \mathrm{~cm}$ in front of the lens.
$M_{T}=\frac{y_{i}}{y_{o}}=-\frac{s_{i}}{s_{o}}=-\frac{-10.8 \mathrm{~cm}}{27 \mathrm{~cm}}=0.40 \quad$ and so $\quad y_{i}=y_{o} M_{T}=(0.40)(9.0 \mathrm{~cm})=3.6 \mathrm{~cm}$
When $M_{T}>0$, the image is upright, and the same conclusion follows from the fact that $y_{i}>0 .$

Key Concepts

Thin Lens Equation

The thin lens equation, 1/s? + 1/s? = 1/f, is a fundamental relation in geometric optics that connects the object distance (s?), the image distance (s?), and the focal length (f) of a lens. It allows us to calculate the position of the image for any given object placement relative to a thin lens.

Virtual Image Formation

Virtual images are formed when the diverging rays of light, after passing through a lens, appear to originate from a point behind the lens. Unlike real images, the light does not actually converge at the image location, making it impossible to project on a screen, and virtual images are typically upright.

Magnification

Magnification in lens systems describes the ratio of the image height to the object height and can also be expressed in terms of the distances of the object and image from the lens. It indicates how the size of the image compares to the object, with the sign of the magnification telling us whether the image is inverted or upright.

Ray Diagrams

Ray diagrams are graphical methods used in optics to determine the position and characteristics of images formed by lenses. By tracing specific rays, such as those parallel to the optical axis or through the optical center, one can accurately construct the location, orientation, and size of the image.

Principal Focus

The principal focus (or focal point) of a lens is the point at which parallel rays of light either converge (in converging lenses) or appear to diverge from (in diverging lenses) after passing through the lens. It plays a critical role in defining the focal length and the overall behavior of the lens in imaging applications.

Optical Center Behavior

The optical center of a lens is the point on the lens through which light rays pass without deviation. This property simplifies image construction since a ray drawn through the optical center will continue in a straight line, aiding in the determination of the image location in ray diagrams.

Transcript

00:01 So as you can see from here by construction we are getting the image i i prime of the object o o prime.

00:07 We have taken a ray which is parallel to the principal axis and after refraction from the lens it seems it is coming from the point focus of the lens.

00:19 And another ray which is going through the optical center, sorry optical center of the lens.

00:27 If these two rays are extended backwards, they will make.

00:31 Meet at this point i so i will be the virtual image of the point o okay so this is the formation of image by construction now from here we can see that the image is formed as you can see in this case we have taken the scale as scale as two centimeter as one centimeter that means two centimeter of the that means two sorry it is better to write like this one centimeter of the drawing scale is equal to two centimeter one centimeter of the drawing scale that is the scale we are using for drawing is equal to two centimeter so from here you can see that the object is actually in 27 centimeter from the lens this is the lens this is the lens so this is 13 .5, that means 27 .5.

01:57 So the image is actually formed.

01:59 You can see from here, from 20, this is 14 .5.

02:03 So image is actually 5 .5, image distance, 5 .5 multiplied by 2.

02:18 That means 11 centimeter.

02:20 Okay, so this is the image distance.

02:22 And the height of the object, as you can see from here, it is, you can see from here that this height is actually taken as 11 cm because it is 10 .5 to sorry 9 cm it is 10 .5 to 15 7 .4 .5 multiplied by 2 that is 9 centimeter and the object height is taken is coming here as actually it is coming 15 and this is you can see this is 16 this is not 17 i think approximately you can say it is actually in the scale of drawing scale it is actually one, not actually, i mean one, two, not actually two, it is less than two.

03:19 I think approximately you can say that this is actually 1 .9 of the drawing scale, okay, or 1 .8.

03:29 So from here we can write that the image height, height of the image or image height is equal to actually we can see it is actually like 1 .8.

03:54 So 1 .8 multiplied by 2, that means 3 .6 centimeters.

03:58 There are some error due to that drawing, i mean that any drawing is not perfect.

04:04 So we are drawing it like this.

04:06 So this is the thing...

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