Home Page Textbooks Nelson Science Perspectives 10 Page 584: Self-Quiz

Nelson Science Perspectives 10

1st Edition

Christy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe

ISBN: 9780176355289

Textbook solutions

All Solutions

Page 584: Self-Quiz

Exercise 11

Solution 1

Solution 2

Step 1

1 of 1

It is possible for a converging lens to not to produce a real image of an object placed in front of it. This is done by placing an object within its focal length. The image produced by this location of the object is virtual in nature and upright in attitude and not real.

Step 1

1 of 2

It is possible for a converging lens not to produce real image of an object which is placed in front of it when we place the object within its focal length.

Produced image will be virtual and upright.

Result

2 of 2

It is possible.

Exercise 12

Step 1

1 of 3

Data:

$$
f=15
$$

$$
d_o=32
$$

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{15(32)}{32-15}=dfrac{480}{17}
$$

Evaluate:

$$
d_i=28.235
$$

The image formed is $28.235$ centimeters away from the lens on its other side as the object.

Result

3 of 3

The image formed is $28.235$ centimeters away from the lens on its other side as the object.

Exercise 13

Step 1

1 of 2

a. Yes the image is produced on the same side of the lens as the figurine. This is because all virtual images are always formed on the same side of the lens as the object.

Step 2

2 of 2

Data:

$$
M=2.9
$$

$$
d_o=6.5
$$

Magnification equation:

$$
M=dfrac{d_i}{d_0}
$$

Rearrange to solve for $d_i$:

$$
d_i=Mtimes d_0
$$

Substitute values:

$$
d_i=2.9times 6.5
$$

Evaluate:

$$
d_i=18.85
$$

The image formed is $18.85$ centimeters away from the lens on its same side as the object.

Exercise 14

Step 1

1 of 3

Data:

$$
h_o=10.4
$$

$$
h_i=4.1
$$

Step 2

2 of 3

Magnification equation:

$$
M=dfrac{h_i}{h_0}
$$

Substitute values:

$$
M=dfrac{4.1}{10.4}
$$

Evaluate:

$$
M=0.394
$$

The magnification of this diverging lens is 0.394

Result

3 of 3

The magnification of this diverging lens is 0.394

Exercise 15

Step 1

1 of 3

Data:

$$
f=22
$$

$$
d_o=2(63)=126
$$

Step 2

2 of 3

Lens equation:

$$
dfrac{1}{d_0}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Rearrange to solve for $d_i$:

$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_0}
$$

Simplify using least common denominator:

$$
dfrac{1}{d_i}=dfrac{d_0-f}{fd_0}
$$

Rearrange:

$$
d_i=dfrac{fd_0}{d_0-f}
$$

Substitute values:

$$
d_i=dfrac{22(126)}{126-22}=dfrac{2772}{104}
$$

Evaluate:

$$
d_i=26.65ne2(34)=68
$$

When the distance of the object from the lens is doubled, the distance between the image does not double. It actually decreases from 34 centimeters to 26.65 centimeters.

Result

3 of 3

No.

Exercise 16

Solution 1

Solution 2

Step 1

1 of 2

$textit{a.}$, Film in the movie projector is loaded into the projector upside-down.

$textit{b.}$, It is inverted so that the image which will be formed on the screen, which will also be inverted by the lens, can be upright.

Result

2 of 2

$textit{a.}$, It is upside-down.

$textit{b.}$, Because it will later be inverted by the lens so it will appear upright on the screen.

Step 1

1 of 1

A image formed by a projector is real in nature and therefore inverted with respect to the orientation of the object. On the screen of the projector however an upright image is required and in order to make this happen, the film is loaded upside down. Its real image formed on the screen will therefore be inverted image of the upside film and so upright in attitude.

Exercise 17

Step 1

1 of 2

a. Presbyopia is a form of far-sightedness in which the images of the objects near the eyes are blurry. These blurred images are formed because the eye muscles have lost their ability to adjust the eye lens to focus a sharp image on the retina. This condition usually occurs in old age as the muscles age with time. By holding the book further away, such people are better able to see and read the sentences in the book.

Step 2

2 of 2

b. A positive meniscus lens or a converging lens can be used to correct this condition. This is possible because the corrective lens converges the light rays so that a sharp image is formed on the screen. Diagram: Exercise scan

Exercise 18

Solution 1

Solution 2

Step 1

1 of 1

A negative distance of image implies that the image formed is virtual in nature and because of this it is on the same side of the lens as the object. Therefore this distance is taken to be negative, as a sign convention positive sign is associated with real image while negative for virtual image.

Step 1

1 of 2

Distance is taken to be negative since the negative distance of an image implies that the image which is formed is virtual and on the same side of the lens as the actual object.

(Negative sign is associated with virtual image, and positive with the real one by the sign convention)

Result

2 of 2

Sign is negative because image which is formed is virtual.

Exercise 19

Solution 1

Solution 2

Step 1

1 of 1

The lens of a camera is a converging lens. As the object distance between the lens and the object increases, the distance between the lens and the image decreases and as a result of which the image formed is smaller in size than the object. The same has happened here, by moving farther away, the object distance has increased, decreasing the image distance and its size.

Step 1

1 of 2

What happened here is that the distance of the object has increased which made image distance and size decrease.

We can explain this by noticing that camera has a converging lens which lead to conclusion that when distance between lens and object increases, distance between lens and image will decreases which will result as image which is smaller in size than the actual object.

Result

2 of 2

In this case, distance of the object has increased which made image distance and size decrease.

Exercise 20

Step 1

1 of 1

WINDOW TO A NEW WORLD HAS BEEN CREATED!

Antony van Leeuwenhoek, businessman and self-taught scientist has presented new discovery today!

A simple device, a powerful magnifying glass which is using one lens and the tiny hole is mounted in the brass plate. There is a specimen which is mounted on the point which is sticking up in front of the lens and by turning two screws focus can be changed.

When you look through the tiny hole, you have to be patient, and can see a completely new world, which we cannot see just by observing with our eyes. There are, how he calls them, “bacteria, parasites”, even the “cells” of our blood could be seen thanks for this magnificent device.

Now that we are capable of a new way if observation, we may get to know the world which surrounds us much better.

Exercise 21

Solution 1

Solution 2

Step 1

1 of 2

Magnification in terms of height is ratio of the image height and the object height and magnification in terms of distance is the ratio of the distance between the image and lens and the distance between the lens and the object.

Result

2 of 2

Step 1

1 of 3

Magnification in terms of height is the ratio of the image height and the object height.

Step 2

2 of 3

Magnification in terms of distance is the ratio of the distance between the image and the lens and the distance between the lens and the object.

Result

3 of 3

Click to see answers.

Exercise 22

Solution 1

Solution 2

Step 1

1 of 2

b. A refractive telescope will be appropriate for an astronomer to observe a star in a distant galaxy because refractive telescope are designed to view objects that are an infinite distance from the earth.

Step 2

2 of 2

a. A terrestrial telescope will be appropriate for a sailor on open water to view a distant land mass because terrestrial telescopes are a modified version of refractive telescope and contain a third lens to produce an upright virtual image of distance objects of earth.

Step 1

1 of 2

$textit{a.}$, A terrestrial telescope will be appropriate for a sailor on an open water to view a distant landmass since they are modified version of refractive telescopes and contain third lens which produces an upright virtual image of distance objects.

$textit{b.}$, A refractive telescope will be appropriate for an astronomer to observe a star in distant galaxy since refractive telescopes are designed for observing objects which are in infinite distance from earth.

Result

2 of 2

$textit{a.}$, A terrestrial telescope.

$textit{b.}$, A refractive telescope.

Exercise 23

Solution 1

Solution 2

Step 1

1 of 2

Part of the camera which has function similar to the function of retina in human eye is photographic film or charged couple device on which image of the actual object is formed.

Result

2 of 2

Photographic film or charged couple device on which image of the actual object is formed.

Step 1

1 of 3

The equivalent of the eye lens in humans is the camera lens.

Step 2

2 of 3

The equivalent of the retina in humans is the photographic film or a charged couple device in cameras on which the image of the object is formed.

Step 3

3 of 3

The equivalent of the focusing of the eye lens to produce share images in human eyes is the focusing of the camera lens. Focusing in camera can also be done manually.

Exercise 24

Solution 1

Solution 2

Step 1

1 of 1

A beam of light entering a converging lens from the left side becomes focused on to the focus of the lens on the right side of the lens, where by a beam of light entering a converging lens from the right side becomes focused on to the focus of the lens on the left side of the lens. This implies that each lens has two principal foci on its either sides. They are both equidistant away from the lens.

Step 1

1 of 2

Each lens has two principal focuses on each side and they are both equidistant from the centre of the lens.

This can be explained by this: Incident ray which is entering the converging lens from the left side becomes focused in one point, focus, on the right side of the lens and vice versa.

Result

2 of 2

We can explain that by: When the incident ray which is entering the converging lens from the left side becomes focused in one point, focus, on the right side of the lens and vice versa.

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