an object is placed at a distance of 50 cm from a concave lens of focal length 20 cm find the nature and the position of the image<br /><br />please help me to solve this problem<br />no spam allowed <br />other wise I will report your answer
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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.
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