Question

A lens forms an image of an object. The object is 16cm from the lens. The image is 12 cm from the lens on the same side as the object. 1) what is the focal length of the lens 2) is the lens converging or diverging 3)if the object is 8.50mm tall how tall is the image 4) is it erect or inverted

Answer 1

1.)

From lens formula,

1/f = 1/v - 1/u

here, v = distance of image from lens = -12 cm (since image is on same side so it's virtual)

u = distance of object from lens = 16 cm

So, 1/f = 1/16 + 1/(-12)

f = focal length = 16*12/(12 - 16)

f = -48 cm = focal length of lens

2.)

Since focal length of converging lens have positive sign.

focal length of diverging lens have negative sign.

from above part,

focal length of this lens have negative sign.

therefore this is a diverging lens.

3.)

magnification of lens is given by,

m = -v/u = -(-12)/16

m = 0.75

also, magnification(m) = hi/ho

here, hi = height of image = ??

ho = height of object = 8.50 mm = 8.50*10^-3 m

So, hi = m*ho = (0.75)*(8.50*10^-3)

hi = 0.006375 = 6.375*10^-3 m

hi = 6.375 mm

4.)

here positive sign means for upright image.

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