e left of the lens is a cube havinQ a race area or 100 cm 2 The base or the cube
ID: 2041191 • Letter: E
Question
e left of the lens is a cube havinQ a race area or 100 cm 2 The base or the cube is on the a Fiq re P36 40 shows a thin qlass ? to the left of the lans 1.54 conver inq lens for hich the adii of curvature are R -15 0 c and R2-11.5 cm. To s or the lens, and the right face is 20.0 on 20.0 cm Figure P36,40 a) Determina the tocal langth ot the lens. 12.06 h Draw tha Image of tha squara faca formed by the lens, hy datermining the following charactaristics of tha limage. tha drawing an paper. Your Instriuctar may ask you to turn it in.) magnitude of height (cm position (cm) 07 eft face our responsc differs from the correct Jnswer by morc than 10%. Double check vour caculator our rc onsc diffcrs from the correct nover by morc than 10%. Doublc check 7.9 7.5 right face our rc onsc differs from the correct answer by morc than 10%. Double check vour ca culatons. Your responsc diffcrs fram the correct answer by morc than 10%. Doublc check your caculat on The oube is imaged in the following two-dimensional geometric shape.Explanation / Answer
given
n = 1.54
R1 = 15 cm
R2 = -11.5 cm
a)
we know, lens makers formula,
1/f = (n - 1)*(1/R1 - 1/R2)
1/f = (1.54 - 1)*(1/15 - 1/(-11.5))
f = 12.05 cm
b) for left face,
u1 = 20 + 10 = 30 cm (object distance)
let v1 is the image distance.
use, 1/u1 + 1/v1 = 1/f
1/v1 = 1/f1 - 1/u1
1/v1 = 1/12.05 - 1/30
v1 = 20.14 cm (position of left face)
height = magnification*h
= (v1/u1)*h
= (20.14/30)*10
= 6.71 cm
Inverted
for right face,
u2 = 20 = 20 cm (object distance)
let v2 is the image distance.
use, 1/u2 + 1/v2 = 1/f
1/v2 = 1/f - 1/u2
1/v2 = 1/12.05 - 1/20
v2 = 30.3 cm (position of right face)
height = magnification*h
= (v2/u2)*h
= (30.3/20)*10
= 15.1 cm
Inverted
c) trapezoid.
d) area = (h2 + h1)*(v2 - v1)/2
= (15.1 + 6.71)*(30.3 - 20.14)/2
= 111 cm^2
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