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An object is 2 cm from a lens which forms an erect image 1/4th (exactly) the size of the object .Determine the focal length of the lens.

What type of lens is this?


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Diminished erect image is formed in case of concave lens. So it is a concave lens.
object distance (u) = -2cm
image distance  = v cm
magnification (m) = +(1/4)

m = v/u
⇒ +1/4 = v/(-2)
⇒ v = -2/4 = -0.5 cm

 \frac{1}{u} + \frac{1}{v} = \frac{1}{f}  \\  \\  \frac{1}{-2} + \frac{1}{-0.5} = \frac{1}{f}  \\  \\ -0.5-2= \frac{1}{f}  \\  \\ \frac{1}{f} =-2.5 \\  \\ f =  \frac{1}{-2.5}=-0.4cm

Focal length is -0.4 cm. It is a concave lens as indicated by the negative sign of focal length.
2 5 2
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