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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?
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Solution
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.
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