Physics, 16.04.2020 19:11 electrofy456

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An object is placed 20 cm away from a curved mirror that has a focal length of 7 cm.
Where is the image located? Is the image magnified or shrunken? Show your work

An image is produced 51.5 cm away from a converging lens. If the object is placed at 32 cm away, where is the focal point located? Show your work

If you used a converging lens with a focal length of 17 cm, where would you put an object that has a height of 5.5 cm to create a virtual image with a height of 13.8 cm? The image is located 22 cm away from the mirror. Show your work

Describe virtual images and how the simulation showed why they are virtual

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Answer from: Ailyt2543

1) 10.8 cm, shrunken

2) 19.7 cm

3) 8.8 cm

4) See explanation

Explanation:

A mirror works on the principle of reflection: a ray of light coming from an object is reflected back by the mirror, producing an image of the object.

The position of the image can be found by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the mirror

p is the position of the object

q is the location of the image

In this problem:

f = 7 cm is the focal length of the mirror

p = 20 cm is the position of the object

So we can find q from the equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{7}-\frac{1}{20}=0.0929\\q=\frac{1}{0.0929}=10.8 cm$

The sign of q is positive: this means the image is real.

The magnification of the image is given by

$M=\frac{y'}{y}=-\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

M is the magnification

Substituting,

$M=-\frac{10.8}{20}=-0.54$

Which means that he image is shrunken, (because |M| ), so .

A lens works on the principle of refraction: the rays of light coming from an object are refracted into the focal point of the lens, and they can produce an image of the object.

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the length

p is the position of the object

q is the location of the image

In this problem we have:

q = 51.5 cm is the location of the image

p = 32 cm is the position of the object

Solving for f, we find the focal length:

$\frac{1}{f}=\frac{1}{32}+\frac{1}{51.5}=0.051\\f=\frac{1}{0.051}=19.7 cm$

In this problem, we know the following data:

f=+17 cm is the focal length (for a converging lens, it is positive)

y=5.5 cm is the height of the object

y'=13.8 cm is the height of the image

q=-22 cm is the location of the image (it is negative because the image is virtual)

From the magnification equation, we can find the position of the object:

$\frac{y'}{y}=-\frac{q}{p}$

And solving for p, the position of the object, we find:

$p=-\frac{qy}{y'}=-\frac{(-22)(5.5)}{13.8}=8.8 cm$

So, the object should be located 8.8 cm from the lens.

The images produced by a lens or by a mirror can be of two types:

Real: a real image is produced when the refracted (or reflected) rays actually meet behind the lens (or in front of the mirror). In this case, the image can be actually projected on a screen. A real image is located behind the lens (or in front of the mirror).Virtual: a virtual image is produced when the refracted (or reflected) rays do not meet at any point, so no image can be actually projected on a screen. In this case, the image forms from the intersection of the prolongation of the refracted/reflected rays. A virtual image is located in front of the lens (or behind the mirror).

Answer from: Quest

c)as speed changes, wavelength and frequency change.

everything changes!

Answer from: Quest

reserve is the answer