Question

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer 1

Case I: (In case of convex mirror)

Magnification, `"m" = 1/3` (convex mirror)

Focal length (f) = 20 cm

Size of image `("h"_"i") = 1/3 xx "h"_"o"` (Size of object)

Using the magnification formula,

`"m" = "h"_"i"/"h"_"o" = (-upsilon)/"u"`

Substituting the values in the above equation,

`("h"_"o"/3)/"h"_"o" = (-upsilon)/"u"`

`1/3 = (-upsilon)/"u"`

`upsilon = (-"u")/3`

Using the mirror formula,

`1/upsilon + 1/"u" = 1/"f"`

`(-3)/"u" + 1/"u" = 1/20`

`(-2)/"u" = 1/20`

u = -40 cm

Negative sign shows that the object is in front of the mirror.

Substituting the above value in obtained image distance and object distance relation,

`upsilon = -u/3 = - ((-40))/3 = 40/3` cm

Thus, the image formed is virtual, erect and diminished. The object is at a distance of 40 cm from the mirror.

Case II: In case of concave mirror

Magnification, `"m" = -1/3` (concave mirror)

Focal length (f) = -20 cm

Using the magnification formula,

`"m" = "h"_"i"/"h"_"o" = (-upsilon)/"u"`

Substituting the values in the above equation,

`(-1)/3 = (-upsilon)/"u"`

`1/3 = upsilon/"u"`

`upsilon = "u"/3`

Using the mirror formula,

`1/upsilon + 1/"u" = 1/"f"`

`3/"u" + 1/"u" = 1/(-20)`

`4/"u" = 1/(-20)`

u = -80 cm

`upsilon = "u"/3 = -80/3` cm

Thus, the image will be real and inverted. The object is at a distance of 80 cm from the mirror.