An object of height 1.2m is placed before a concave mirror of focal length 20cm so that a real image is formed at a distance of 60cm from it. Find the position of an object. What will be the height of the image formed?
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Here is your answer!
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I have taken height of object = 1.2 cm
In question it is given 1.2 m
Object is always place to the left side of a mirror therefore the object distance (u) is always negative.
If the image is formed in front of the mirror(real image) on the left side then the image distance (v) will be negative.
The focus of concave mirror is in front of the mirror on the left side so the focal length of a concave mirror is considered negative.
Given:
Focal length (f)= -20 cm ( concave mirror)
Image distance (v)= -60cm (real image)
Height of the object (h1)= 1.2cm
Mirror Formula
1/v + 1/u = 1/f
(-1/60)+(1/u) = (-1/20)
1/u= -1/20 + 1/60
1/u = (-3+1)/60= -2/60= -1/30
1/u= -1/30
u = -30
Object distance(u) = -30 cm
Magnification (m) = -v/u
m = - (-60) /-30= 60/-30= -2
m = h2/h1
-2= h2/1.2
-2×1.2= h2
h2= -2.4 cm
Height of real image (h2)= -2.4 cm
Real image is formed below the axis. So the height of a real image is negative.
Hence, the position of an object is -30 cm(to the left of mirror) & the height of the real image formed is -2.4 cm
==========================================================
Hope it helps!
Here is your answer!
==========================================================
I have taken height of object = 1.2 cm
In question it is given 1.2 m
Object is always place to the left side of a mirror therefore the object distance (u) is always negative.
If the image is formed in front of the mirror(real image) on the left side then the image distance (v) will be negative.
The focus of concave mirror is in front of the mirror on the left side so the focal length of a concave mirror is considered negative.
Given:
Focal length (f)= -20 cm ( concave mirror)
Image distance (v)= -60cm (real image)
Height of the object (h1)= 1.2cm
Mirror Formula
1/v + 1/u = 1/f
(-1/60)+(1/u) = (-1/20)
1/u= -1/20 + 1/60
1/u = (-3+1)/60= -2/60= -1/30
1/u= -1/30
u = -30
Object distance(u) = -30 cm
Magnification (m) = -v/u
m = - (-60) /-30= 60/-30= -2
m = h2/h1
-2= h2/1.2
-2×1.2= h2
h2= -2.4 cm
Height of real image (h2)= -2.4 cm
Real image is formed below the axis. So the height of a real image is negative.
Hence, the position of an object is -30 cm(to the left of mirror) & the height of the real image formed is -2.4 cm
==========================================================
Hope it helps!
aharon2003:
somewhat
Answered by
10
U=-30 HEIGHT OF IMG=-V/U
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