The image of an object placed at a distance of 30 cm on the principal axis of a concave mirror from its pole, is formed on the object itself. Find (a) the focal length and (b) linear magnification of mirror.
Answers
Answer:
given u=-30
a/c to que.
(i) image formed at object so, object kept at centre of curvature
therefore, f=-30/2=-15cm
(ii) as image formed at center of curvature so height of the image= height of obj.
m= -h2/h1=-v/u
m=-30/30=-1
-ve sign shows that image is inverted
Answer:
The minimum distance of the object and its real image from a concave mirror is zero and this occurs when the distance between the object and the mirror is equal to twice the focal length.
Explanation:
The Lens/Mirror formula is
1/u + 1/v = 1/f
u = +15 and f = +10 (both Real distances, hence +)
1/15 + 1/v = 1/10
1/v = 1/10 - 1/15 = 6/60 - 4/60 = 2/60 = 1/30
v = +30
so a real image 30cm from the mirror.
M = v/u = +2 so the image is twice the size of the object. M>0 so inverted.