As shown an object is placed 40 cm in front of a converging lens that has f=+8.0 cm . A plane

As shown an object is placed 40 cm in front of a converging...

As shown an object is placed $40 \mathrm{~cm}$ in front of a converging lens that has $f=+8.0 \mathrm{~cm} .$ A plane mirror is $30 \mathrm{~cm}$ beyond the lens. Find the positions of all images formed by this system. For the lens $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8.0}-\frac{1}{40}=\frac{4}{40} \quad \text { or } \quad s_{i}=10 \mathrm{~cm} $$ This is image $A^{\prime} B^{\prime}$ in the figure. It is real and inverted. $A^{\prime} B^{\prime}$ acts as an object for the plane mirror, $20 \mathrm{~cm}$ away. A virtual image $C D$ is formed $20 \mathrm{~cm}$ behind the mirror. Light reflected by the mirror appears to come from the image at $C D$. With $C D$ as object, the lens forms an image of it to the left of the lens. The distance $s_{i}$ from the lens to this latter image is given by $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{50}=0.105 \quad \text { or } \quad s_{i}=9.5 \mathrm{~cm} $$ The real images are therefore located $10 \mathrm{~cm}$ to the right of the lens and $9.5 \mathrm{~cm}$ to the left of the lens. (This latter image is upright.) A virtual inverted image is found $20 \mathrm{~cm}$ behind the mirror.

As shown an object is placed $40 \mathrm{~cm}$ in front of a converging lens that has $f=+8.0 \mathrm{~cm} .$ A plane mirror is $30 \mathrm{~cm}$ beyond the lens. Find the positions of all images formed by this system.
For the lens
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8.0}-\frac{1}{40}=\frac{4}{40} \quad \text { or } \quad s_{i}=10 \mathrm{~cm}
$$
This is image $A^{\prime} B^{\prime}$ in the figure. It is real and inverted.
$A^{\prime} B^{\prime}$ acts as an object for the plane mirror, $20 \mathrm{~cm}$ away. A virtual image $C D$ is formed $20 \mathrm{~cm}$ behind the mirror.

Light reflected by the mirror appears to come from the image at $C D$. With $C D$ as object, the lens forms an image of it to the left of the lens. The distance $s_{i}$ from the lens to this latter image is given by
$$
\frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{50}=0.105 \quad \text { or } \quad s_{i}=9.5 \mathrm{~cm}
$$
The real images are therefore located $10 \mathrm{~cm}$ to the right of the lens and $9.5 \mathrm{~cm}$ to the left of the lens. (This latter image is upright.) A virtual inverted image is found $20 \mathrm{~cm}$ behind the mirror.

Key Concepts

Sign Conventions in Geometric Optics

Accurate use of sign conventions is essential in optics, as they define the positive and negative directions for object distances, image distances, and focal lengths. These conventions help in correctly applying formulas like the lens equation and in distinguishing between different types of images based on their orientations and positions relative to the optical elements.

Sequential Imaging in Optical Systems

When multiple optical elements, such as a lens and a plane mirror, are used in sequence, the image formed by one element can serve as the object for the next. Understanding the cumulative effect of successive imaging processes is essential in designing and analyzing complex optical systems, as it determines the final location and characteristics of the image.

Lens Formula

The lens formula, given by 1/f = 1/s? + 1/s?, is fundamental in analyzing how lenses form images. It relates the distances from the lens to the object and to the image (s? and s? respectively) to the focal length (f) of the lens. This relationship is crucial to predicting where an image will be located and determining if the image is real or virtual.

Real vs Virtual Images

In geometrical optics, images formed by lenses or mirrors are classified as real or virtual. Real images are formed when light rays physically converge and can be projected onto a screen, whereas virtual images appear to originate from a location where light rays do not actually converge, though they can be seen by an observer looking into the optical system.

Plane Mirror Reflection

A plane mirror creates an image through the law of reflection, where the angle of incidence equals the angle of reflection. The image produced by a plane mirror is virtual, upright, and of the same size as the object, appearing to be the same distance behind the mirror as the object is in front.

Transcript

00:01 Hi, in this given problem there is a converging lens first of all and this converging lens has been kept in front of a plain mirror like this as shown in the figure.

00:35 The gap between this converging lens and the plane mirror that is given as 30 centimeters.

00:44 Focal length of this converging lens is given as f is equal to plus 8 .0 centimeter and an object has been kept in front of this converging lens at a distance object distance so is equal to plus 40 centimeter now we have to find the distances of all the images formed by this combination.

01:17 So first of all using lens equation which is a relation among the image distance si the object distance so and the focal length of the converging lens and plugging in all known values here one by si we have to find it plus one by so 40 is equal to 1 by 8.

01:48 So this 1 by si is equal to 1 by 8 minus 1 by 40.

01:55 Lcm here will be 40.

01:58 So this is 5 minus 1 is equal to 4 by 40.

02:04 So si the distance of image comes out to be plus 10 centimeters so that is towards right of the lens means it is somewhere here the image the primary image the first image so its distance will be 10 centimeter now this image will serve as an object for the plane mirror at a gap of 20 centimeter so using the rules for the image formation by a plane mirror the image will be formed at the same distance behind the mirror.

02:47 So this is i1 the first image it will be i2 the second image.

02:56 The distance of first image this is the first answer for this given problem now this image will serve as a virtual object for the plane mirror from which its distance is the total distance was 30 minus 10 centimeter from the converging lens so that remaining distance is 20 centimeter so the image formed by this plain mirror will be behind the mirror at the same distance so, the distance of second image that is s .i .2.

04:38 So we can mark this s .i .1...

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