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Problem 6.88:A concave mirror forms an image, on a screen $2.00 \:m$ in front of the mirror, of an object $20.0 \:cm$ in front of the mirror. What is the image height if the object height is 5.00 mm?

Solution:

Here the object  is placed at a distance of $20.0 \:cm$ from the mirror and the image is formed on the screen at  a distance of $3.00 \:m$. Both image and object are on the same side of the mirror.

Object distance, $u\:=\:-20\:cm$ and image distance $v\:=\:-2\:m\:=\:-200\:cm$.

Lateral magnification is the ra tio of the size of the image to the size of the object.

Or, $m\:=\frac{h’}{h}$ Where $h$ is the size or height of the object and $h’$ is the height of the image.

Lateral magnification is the ratio of the image distance to the  object distance.

Or, $m\:=\:\frac{-image\:distance}{object\:distance}\:=\:\frac{-v}{u}\:=\:\frac{-(-200)}{-20}\:=\:-10$.

The image will be ten times larger than the object. Negative sign indicates that the image is inverted with respect to the object.

Image size $h’\:=\:m\times h\:=\:-10\times 5\:cm\:=\:-50\:cm$.

So the mirror forms an inverted image of size of $50\:cm$.