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An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? |
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Answer» Solution :For CONCAVE lens, Here object height h =3 cm Object distance U = - 14 cm Focal length f = - 21 cm Image distance v = ?  `rArr` Lens formula, `1/f=1/v - 1/u ` `therefore1/v = 1/f + 1/u = 1/(-21) + (1)/(-14)` `therefore 1/(v) = -(1)/(21) - (1)/(14) = (-2-3)/(42) = (-5)/(42)` `therefore v = - (42)/(5) cm = - 8.4 cm ` Negative sigh represent that image is virtual erect and siminised and toward object Maginification `m = v/u ` `therefore h./h = (-8.4)/(-14)` `thereforeh. = 0.6 xx h = 0.6 xx 3` `therefore` Height of image = 1.8 cm As the object GOES away from lens , the vritual image goes toward the optica centre oflens (but it is never obtained away from optical centre) but it is obtained of dereasing size. |
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