A converging lens has a focal length of f= 16.0 cm. An object is placed p_1 = 48

Question

A converging lens has a focal length of f= 16.0 cm. An object is placed p_1 = 48.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. An object is placed p_2 = 16.0 cm from the lens. Find the image distance and describe the image. An object is placed p_3 = 8.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. Suppose the image of an object is upright and magnified 1.55 times when the object is placed 15.5 cm from a particular converging lens. Find the location of the image. Find the focal length of the lens.

Explanation / Answer

(a)

Using lens formula,
object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/do + 1/di
1/di =1/do -1/f
=1/48-1/16
=-24.03
negative virtual upright images

M = hi/ho = -di/do=-(-24.08)/48
=0.0500

We get a positive value to indicate that the object is not inverted.

(b)

Using lens formula,
object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/do + 1/di
1/di =1/do -1/f
=1/16-1/16
=-0
erect image upright images

M = hi/ho = -di/do=-(-0)/48
=0
object is formed at infinty

(c)

Using lens formula,
object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/do + 1/di
1/di =1/do -1/f
=1/8-1/16
=16
real image

M = hi/ho = -di/do=-(16)/8
=-2
object is formed erect

(a)image location is 1.55 cm ehind the lens

(b)1/f = 1/do + 1/di

=1/15.5+1/1.55
f=1.40cm
image is behind the lens

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