# 10 how far from the second lens is the final image of an object infinitely far from the first lens? Ideas

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Ch 26 Homework

## Chapter 26: Optical Instruments

Ch 26: 2,
3, 5, 9, 13, 15, 20, 25, 27

|

26.2 A pin-hole camera is used to take
a photograph of a student who is 1.8 m tall. The student stands
2.7 m in front of the camera. Film is placed 10 cm behind the
pin-hole. Determine the height of the image
produced.

The triangle formed by the object (the
student) and the rays from the student’s head and feet going to
the pin-hole and the triangle formed by the image and the rays
coming from the pin-hole toward the head and feet of the image are
similar triangles so the ratios of corresponding sides of these
triangles are equal. That is,

di = 0.067 m = 6.7
cm

26.3 The distance from the cornea to
the retina of some particular eye is 2.1 cm. Find the effective
focal length of this eye for an object located

(a) at infinity,

(b) for an object located 10 m
away, and

(c) for an object located 25 cm
away.

We can directly apply the Image
Equation

The image distance, di, is the distance
from the cornea to the retina, 2.1 cm.

For (a), the object distance,
do, is infinite,

f = 2.1 cm

For (b), the object distance is
do = 10 m

Watch the
units!
You can not add meters and
centimeters; you can not add inverse meters and inverse
centimeters. The two terms on the right must have the same units

f = 2.096 cm
2.1 cm

For (c), the object distance is
do = 25 cm

f = 1.94 cm

26.5 An object sits 35 cm to the left
of a converging lens with focal length of 20 cm. 75 cm to the
right is a second converging lens with focal length of 15 cm.
Locate and characterize the final image.

A ray diagram is always a good way
to start. There is usually more real understanding available in
the ray diagram than in “merely” solving the problem numerically.
In addition, a good diagram also provides a place to keep all the
dimensions of the problem.

For the first lens, we begin with the
Image Equation,

or

And we know f1 = 20 cm and
do1 = 35 cm,

di1 = 46.7
cm

This image (which is real, inverted, and
enlarged) becomes the object for the second lens as we again apply
the Image Equation,

or

Since the lenses are 75 cm apart, the
image formed 46.7 cm to the right of lens #1 is located
28.3 cm to the left of lens #2 so the object distance, then,
is

do2 = 28.3 cm

di2 = 31.9
cm

That is, the final image is located 31.9
cm to the right of the second lens. Since di2 > 0
(that is, since di2 = +), we know this image is
real. It is inverted, compared to the object for lens #2.
But that was an inverted image of the original object. Therefore,
this final image is upright.

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The total magnification is the product of
the magnifications of the two lenses,

Mtot = M1
M2

Mtot = (1.33)(1.13)

Mtot = 1.5

26.9 A patient’s eye can focus only on
objects beyond 100 cm. What word characterizes this type of vision
problem? What are the focal length and power of the contact lens
needed to correct this problem?

This patient is farsighted or hyperopic.
This patient has trouble reading a book at a comfortable distance
— such as 25 cm — so we must find a lens that will take an
object at 25 cm (do = 25 cm) and create an image (a
virtual image) at 100 cm (di = – 100 cm).

A diagram may help to clearly establish
that we need

do = 25 cm

di = – 100
cm

It is important to remember or realize
that the image formed is a virtual image. It is on the left side
of the lens in the diagram above. This means the image distance is
negative. Once we know do and di then it is easy to find the focal
length of the lens from the Image Equation,

f = 33.3 cm

To describe the lens in terms of “power”
(in units of diopters) we need to take the inverse of its
focal length in meters,

f = 0.3 m

P = 1/f = 3 diopters

26.10 A certain eye can focus only on
objects closer than 50.0 cm. What word characterizes this
type of vision problem? What sort of contact lens (described both
in focal length and power) will correct this
problem?

This eye is myopic or nearsighted. We
need to find a lens that will take an object at infinity
(do = )
and produce a virtual image at 50.0 cm (di = – 50.0 cm
= – 0.50 m).

Now we know the image and object
distances —

do =

di = – 50.0
cm

It is important to remember or realize
that the image formed is a virtual image. It is on the left side
of the lens in the diagram above. This means the image distance is
negative. Once we know do and di then it is
easy to find the focal length of the lens from the Image
Equation,

f = – 50.0 cm

f = – 0.50 m

P = 1/f = – 2
diopters

Note: This problems started out in
the book with a far point of 500 cm or 5.0 m. As we will see, that
does not require very much correction to be able to see something
at infinity. This does not seem like a reasonable or interesting
problem. Therefore, I changed the problem. However, let’s go ahead
and solve this for the original case —

do =

di = – 500 cm

f = – 500 cm = – 5 m

P = 0.20 diopters

26.11 A particular eye can focus only
on objects more distant than 150 cm. What word characterizes this
type of vision problem? What are the focal length and power of the
corrective lens needed to correct this problem? The lens is to be
worn 2.0 cm in front of the eye?

Using a lens that is 2 cm from the eye is
but a slight variation from problem 26.9 that used a contact lens.
For this situation we have

do = 23 cm

di = – 148
cm

It is important to remember or realize
that the image formed is a virtual image. It is on the left side
of the lens in the diagram above. This means the image distance is
negative. Once we know do and di then it is easy to find the focal
length of the lens from the Image Equation,

f = 27.2 cm = 0.272 m

P = 1/f = 3.7
diopters

26.20 A diamond is viewed with a
jeweler’s loupe that has a focal length of 5.0 cm. Where must the
diamond be placed to provide a virtual image at infinity? What
will be the angular magnification of this simple
magnifier?

The diamond should be placed at the focal
point of the lens (the jeweler’s loupe), that is, 5.0 cm from the
lens.

The magnification is given by

M = 5

26.25 A microscope has its objective
and eyepiece 18 cm apart. If fobj = 0.40 cm and
feye = 5.0 cm, where must a specimem be located to
produce a final virtual image at infinity? What is the total
magnification of this microscope?

To have the final image at di
= , we need the real image formed by the objective to be located
at the focal point of the eyepiece. With the objective and
eyepiece lenses 18.0 cm apart this means,

di + feye =
18.0 cm

di + 5.0 cm = 18.0
cm

di = 13.0
cm

Now we can again apply the Image
Equation,

do = 0.41 cm

26.27 An astronomical telescope is
used to view an object at infinity. The objective lens has a focal
length of 15.0 cm. Where must the 0.5 cm eyepiece be placed to
form an image at infinity? What is the total angular
magnification?

The real image formed by the objective
lens will be located at the focal point for that lens,
di = fobj = 15.0 cm from that lens. To
produce a virtual image at infinity, the eyepiece must be located
so that this image — which acts as an object for the eyepiece —
is located at the focal point for the eyepiece lens,
do = feye = 0.5
cm.

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Therefore, the eyepiece must be 15.5 cm
from the objective lens.

The total magnification is given
by

M = 30

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## Frequently Asked Questions About how far from the second lens is the final image of an object infinitely far from the first lens?

If you have questions that need to be answered about the topic how far from the second lens is the final image of an object infinitely far from the first lens?, then this section may help you solve it.

### How do you figure out how far an image is from the lens?

Determine whether the focal length is positive or negative in step one. Determine whether the image distance is positive or negative in step two. Calculate the distance of the object from the lens using the formula do=11f?1di d o = 1 1 f? 1 d i in step three.

### The image’s distance from the lens will be equal to when the object is at infinity.

As a result, for all of the lenses in this scenario, the image distance will always be equal to the focal lengths of the lenses, i.e., the image distance for all of the lenses in this scenario is equal to the focal length of the lenses.

### What transpires when an object is infinitely distant?

Therefore, even when the object is infinitely distant, an image will still form, and the image’s location (when the object is far away) occurs at the focal length of the lens.

### How do you calculate the separation between an object and its image?

The graph between object distance u and image distance v for a lens is shown below. The focal length of the lens is. According to lens formual, we have where u = distance of the object from the lens, v = distance of the image from the lens, and f = focal length of given lens.

### What is the image distance formula?

The image will always be behind the object, on its side of the lens, and somewhere farther away from the lens, regardless of precisely where the object is in front of F.

### When an object is far from the lens, where is the image?

When an object is placed at infinity, a virtual image is formed at focus and the size of the image is smaller. An image formed by a convex mirror is always virtual and erect.

### When an object is positioned far from a convex mirror, where will the image be formed?

The image formed is at the focus of the mirror if the object is at infinity from the concave mirror because parallel light rays that are reflected converge at the focus.

### If an object is positioned at an infinite distance from a concave mirror B at its focal point, where will the image be formed?

Q. Q. When an object is placed at infinity, its image is formed at the focus behind the convex mirror.

### What is the image distance formula?

You can find the distance between two points by using the one-dimensional distance formula, which is d = |x2 – x1|. Usually, you’ll want to use the one-dimensional distance formula when your two points lie on a number line or axis.

### How do you figure out how far away something is?

The focal length is a property of the thin lens, and it specifies the distance at which parallel rays come to a focus. The image distance is simply the distance from the object to the thin lines.

### What is a lens’ image distance?

The lens formula is used to calculate magnification. It is given in terms of image distance and object distance, and is equal to the ratio of image distance to that of object distance. The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

### a segment from the YouTube video How to get even spacing/Equal distance/Calculate

Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t. To solve for distance, use the formula for distance d = st, or distance equals speed times time.

### How do you figure out how far you’ve travelled?

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The distance formulas for points in rectangular coordinates in two- and three-dimensional Euclidean space are based on the Pythagorean theorem. For example, the distance between the points (a,b) and (c,d) is given by Square root of (a? c)2 + (b? d)2.

### a portion of the YouTube video Distance Formula – How to Use

The Pythagorean Theorem, which states that a side plus a side equals a side equals a side, or a side plus a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals a side equals