What Type of Mirror Creates an Inverted, Real Image Half the Object's Size?

[Phower Power]

An object is placed 11 cm from a certain mirror. The image is half the height of the object, inverted, and real.


1/p+1/i=1/f



? Clueless because I don't know how to set up the equation int his kind of situation.

 

[tiny-tim]

Welcome to PF!

Hi Phower Power! Welcome to PF!

… I don't know how to set up the equation int his kind of situation.

I don't get it …

I assume you do know how to set it up in some other situation?

Just do the same, and convert "The image is half the height of the object, inverted, and real" into an equation also.

 

[technician]

do you know the equation for magnification?

 

[Phower Power]

Nm, I got it! =)

Ho/Hi = magnification

Then I plug it in! Thank you for the clues

 

[asteriatic]

I can provide a response to this content by explaining the concepts of reflection and refraction and how they relate to the given situation. Reflection is the bouncing back of light rays when they hit a surface, while refraction is the bending of light as it passes through a medium with a different density.

In this scenario, an object is placed 11 cm away from a mirror. The image produced by the mirror is half the height of the object, inverted, and real. This means that the image is the same size as the object, but appears upside down and can be projected onto a screen. This type of image is called a real image, as opposed to a virtual image which cannot be projected.

To understand the relationship between the object, image, and mirror, we can use the mirror equation: 1/p + 1/i = 1/f, where p is the distance of the object from the mirror, i is the distance of the image from the mirror, and f is the focal length of the mirror. The focal length is the distance from the mirror to its focal point, where parallel light rays converge after being reflected.

In this case, we know that the image is half the height of the object, so i = 1/2p. We also know that the image is real, which means it is on the same side of the mirror as the object. This means that i and p have the same sign in the equation. Substituting these values into the mirror equation, we get 1/p + 2/p = 1/f. Solving for f, we get f = 2p.

Therefore, the focal length of the mirror is equal to twice the distance of the object from the mirror. This information can be useful in determining the size and position of the image produced by the mirror, as well as the magnification of the image. By understanding the principles of reflection and refraction, we can apply mathematical equations to analyze and explain various optical phenomena.