an object is placed at a distance of 50 cm of convex lens of power 4. find nature position and magnification
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Object distance u = -50 cm
Focal length of concave lens f = -20 cm
Lens formula: 1/v - 1/u = 1/f 1/v
= 1/f + 1/u = 1/(-20) + 1/(-50)
= -7/100
image distance v = -100/7 = - 14.3 cm
Magnification m = v/u = - 14.3/(-50) = 0.286
The image is virtual, erect, diminished and is formed at a distance of 14.3 cm from the optic centre on the same side of the lens as the object is placed.
Hope its Help you
Focal length of concave lens f = -20 cm
Lens formula: 1/v - 1/u = 1/f 1/v
= 1/f + 1/u = 1/(-20) + 1/(-50)
= -7/100
image distance v = -100/7 = - 14.3 cm
Magnification m = v/u = - 14.3/(-50) = 0.286
The image is virtual, erect, diminished and is formed at a distance of 14.3 cm from the optic centre on the same side of the lens as the object is placed.
Hope its Help you
pihoo747:
how is focal lenght -20?
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