$[1\mathrm{.}5/6$ Points$]$

Note: Make sure to develop some intuition about what the answer will be like, before plugging into the equation.

An object of height $y_0=+2m$ is placed in front of a concave spherical mirror $($focal length of size $0\mathrm{.}5$ m $),$ at the various locations listed in parts a$-$c$.$

a$.)$ When the object is placed $2$ m away from the mirror, please find:

image distance $=$

type$/$orientation of image $=$ real and inverted

magnification $=$

height of image : m

b$.)$ When the object is placed $0\mathrm{.}75$ m away from the mirror, please find:

image distance $=q,$ m

type$/$orientation of image $=$ real and inverted

magnification $=q,$

height of image, $=q,$ m

c$.)$ When the object is placed $0\mathrm{.}25$ m away from the mirror, please find:

image distance $=$ m

type$/$orientation of image $=$ virtual and upright

magnification $=$

height of image $=$ m

The same object $($of height $=2m)$ is placed in front of a converging lens $($focal length of size $0\mathrm{.}5$ m $),$ at the various locations listed in parts d$-$f$.$

d$.)$ When the object is placed $2$ m away from the lens, please find:

image distance $=q,$ m

type$/$orientation of image $=$ real and inverted

magnification $=$

height of image $=$ m

e$.)$ When the object is placed $0\mathrm{.}75$ m away from the lens, please find:

image distance $=$ m

type$/$orientation of image $=$ real and inverted

magnification $=$

height of image $=$ m

f$.)$ When the object is placed $0\mathrm{.}25$ m away from the lens, please find:

image distahie $=$ m

type$/$orientation of image $=$ virtual and upright

magnification $=$

height of image $=$