A diverging lens has a focal length of-8.00 cm. What are the image distances for

Question

A diverging lens has a focal length of-8.00 cm. What are the image distances for objects placed at these distances from the lens: 5.00 cm, 8.00 cm, 14.0 cm, 16.0 cm, 20.0 cm? In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size, If the object is 4.00 cm high, what is the height of the image for the object distances of 5.00 cm and 20.0 cm?. A converging lens has a focal length of 8.00 cm. What are the image distances for objects placed at these distances from the thin lens: 5.00 cm, 14.0 cm, 16.0 cm, 20.0 cm? In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size, If the object is 4.00 cm high, what is the height of the image for the object distances of 5.00 cm and 20.0 cm?. Sketch a ray diagram to show that if an object is placed less than the focal length from a converging lens, the image is virtual and upright, (connect tutorial: magnifying glass). For each of the lenses in the figure, list whether the lens is converging or diverging. In order to read his book. Stephen uses a Pair of

Explanation / Answer

lens maker’s formula is

1/f = 1/v-1/u

f is the focal length of the lens, u is object distance and v is image distance.

we need image distance hence

f= -8.0 cm

v = uf/(u+f)

u = -5.0cm , v= -5*-8.0/(-5-8.0) = -3.08 cm, image virtual, upright, diminished

magnification = v/u = 3.08/5

object height = 4.0 cm

image height = 4*3.08/5 = 2.46 cm

u = -8.0 cm   v = -8.0*-8.0/(-8.0-8.0) = -4.00 cm image virtual, upright, diminished

u= -14.0 cm v=   -14.0*-8.0/(-14.0 – 8.0) = -5.09 cm image virtual, upright, diminished

u = -16.0 cm v = -16.0*-8.0/(-16.0-8.0) = -5.33cm, image is virtual, upright and diminished

u= -20.0cm v = -20.0*-8.0/(-20.0-8.0) = -5.71 cm , image virtual, upright, diminished

magnification = v/u = 5.71/20 = 0.29

object height = 4.0cm

image height = 4.0*0.29 = 1.16 cm

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