Consider an object with = 12 cm that produces an image with 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encou "objects that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges Part A Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation Express your answer in centimeters, as a fraction or to three significant figures. Panor Part Vdo for Panredo foar A reor Part A keyboard shortcuts for Part A help for Part A f= Submit Part B Complete previous part(s) Part C Complete previous part(s) Part D Complete previous part(s) Now consider a diverging lens with focal length f-15 cm, producing an upright image that is 5/9 as tall as the object. Part E Request Answer Is the image real or virtual? Think about the magnification and how it relates to the sign of s @real virtual ▾ Part F Previous Answers Correct cm What is the object distance? You will need to use the magnification equation to find a relationship between aand s'. Then substitute into the thin lens equation to solve for a Express your answer in centimeters, as a fraction or to three significant figures.
Ray Optics
Optics is the study of light in the field of physics. It refers to the study and properties of light. Optical phenomena can be classified into three categories: ray optics, wave optics, and quantum optics. Geometrical optics, also known as ray optics, is an optics model that explains light propagation using rays. In an optical device, a ray is a direction along which light energy is transmitted from one point to another. Geometric optics assumes that waves (rays) move in straight lines before they reach a surface. When a ray collides with a surface, it can bounce back (reflect) or bend (refract), but it continues in a straight line. The laws of reflection and refraction are the fundamental laws of geometrical optics. Light is an electromagnetic wave with a wavelength that falls within the visible spectrum.
Converging Lens
Converging lens, also known as a convex lens, is thinner at the upper and lower edges and thicker at the center. The edges are curved outwards. This lens can converge a beam of parallel rays of light that is coming from outside and focus it on a point on the other side of the lens.
Plano-Convex Lens
To understand the topic well we will first break down the name of the topic, ‘Plano Convex lens’ into three separate words and look at them individually.
Lateral Magnification
In very simple terms, the same object can be viewed in enlarged versions of itself, which we call magnification. To rephrase, magnification is the ability to enlarge the image of an object without physically altering its dimensions and structure. This process is mainly done to get an even more detailed view of the object by scaling up the image. A lot of daily life examples for this can be the use of magnifying glasses, projectors, and microscopes in laboratories. This plays a vital role in the fields of research and development and to some extent even our daily lives; our daily activity of magnifying images and texts on our mobile screen for a better look is nothing other than magnification.

 alt="+ Understanding Lenses
Learning Goal:
To learn the quantitative use of the lens equation, as well as how to determine qualitative
properties of solutions.
In working with lenses, there are three important quantities to consider. The object distance
8 is the distance along the axis of the lens to the object. The image distance s' is the
distance along the axis of the lens to the image. The focal length f is an intrinsic property
of the lens. These three quantities are related through the equation
= + ===
Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a
lens problem always assumes that you are using thin, spherical lenses.
The equation above allows you to calculate the locations of images and objects. Frequently,
you will also be interested in the size of the image or object, particularly if you are
considering a magnifying glass or microscope. The ratio of the size of an image to the size
of the object is called the magnification. It is given by
4
V
m =
where y' is the height of the image and y is the height of the object. The second equality
allows you to find the size of the image (or object) with the information provided by the thin
lens equation.
All of the quantities in the above equations can take both positive and negative values.
Positive distances correspond to real images or objects, while negative distances
correspond to virtual images or objects. Positive heights correspond to upright images or
objects, while negative heights correspond to inverted images or objects. The following
table summarizes these properties:
positive negative
8 real
virtual
real
virtual
y upright
inverted
y' upright inverted
The focal length f can also be positive or negative. A positive focal length corresponds to a
converging lens, while a negative focal length corresponds to a diverging lens.
▾ Part A
Consider an object with a = 12 cm that produces an image with 8 = 15 cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter
"objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.
Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.
Express your answer in centimeters, as a fraction or to three significant figures.
Painfo Far do for Part redo foart A reor Part A keyboard shortcuts for Part A help for Part A
f=
Submit
Part B Complete previous part(s)
Part C Complete previous part(s)
Part D Complete previous part(s)
Now consider a diverging lens with focal length f= -15 cm, producing an upright image that is 5/9 as tall as the object.
▾ Part E
Is the image real or virtual? Think about the magnification and how it relates to the sign of s'
real
Request Answer
virtual
Submit
▾ Part F
Previous Answere
✔Correct.
cm
What is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for 8.
Express your answer in centimeters, as a fraction or to three significant figures.
Pearson
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