Find the position, nature and size of the image formed by a convex lens of focal
length 12 cm of an object 5 cm high placed at a distance 20 cm from it.
AS we are given with :-
Height of Object h = +5 CM
Focal length f = 12 CM
Object distance u = -20 CM
We have to find :-
Image-DIstance v = ?
Height of image h'= ?
AS LENS Formula = 1/v - 1/u = 1/f
OR 1/v = 1/u + 1/f
1/v = 1/-20 + 1/12 gives
1/v = -3 / 60 + 5/60
1/v = 2/60 or 1/v = 1/30
Hence v = +30 Cm
The positive sign of v Shows that the image is formed at a distance of 30 cm on the other side of the optical centre . $$Hence Image is Real and Inverted_$$.
Magnification m = h'/h = v/u or
h' = h(v/u)
Height Of IMage, h'= (5.0) (+30/-20)
h'= 5 x -1.5 = -7.5
$$THe NEgative sign of h' Shows that image is inverted and real _$$
It is formed below the principle axis. Thus, a real,invertedimage,
7.5 cm tall is formed at distance of 30cm on the other side of the lens . The image is ENLARGED
Height of Object h = +5 CM
Focal length f = 12 CM
Object distance u = -20 CM
We have to find :-
Image-DIstance v = ?
Height of image h'= ?
AS LENS Formula = 1/v - 1/u = 1/f
OR 1/v = 1/u + 1/f
1/v = 1/-20 + 1/12 gives
1/v = -3 / 60 + 5/60
1/v = 2/60 or 1/v = 1/30
Hence v = +30 Cm
The positive sign of v Shows that the image is formed at a distance of 30 cm on the other side of the optical centre . $$Hence Image is Real and Inverted_$$.
Magnification m = h'/h = v/u or
h' = h(v/u)
Height Of IMage, h'= (5.0) (+30/-20)
h'= 5 x -1.5 = -7.5
$$THe NEgative sign of h' Shows that image is inverted and real _$$
It is formed below the principle axis. Thus, a real,invertedimage,
7.5 cm tall is formed at distance of 30cm on the other side of the lens . The image is ENLARGED