please answer and show your work, thank you! 1.) A farsighted person has a near

Question

please answer and show your work, thank you!

1.) A farsighted person has a near point that is 48.0 cm from her eyes. She wears eyeglasses that are designed to enable her to read a newspaper held at a distance of 27.0 cm from her eyes. Find the focal length of the eyeglasses, assuming each of the following.

(a) that they are worn 2.0 cm from the eyes
cm

(b) that they are worn 3.0 cm from the eyes
cm

2.) A converging lens (f = 11.4 cm) is located 33.0 cm to the left of a diverging lens (f = -6.06 cm). A postage stamp is placed 39.6 cm to the left of the converging lens.

(a) Locate the final image of the stamp relative to the diverging lens. (Include sign to indicate which side of the lens the image is on.)
____________cm

(b) Find the overall magnification.
________________

3.) The owner of a van installs a rear-window lens that has a focal length of -0.301 m. When the owner looks out through the lens at a person standing directly behind the van, the person appears to be just 0.237 m from the back of the van, and appears to be 0.335 m tall.

(a) How far from the van is the person actually standing?
___________m

(b) How tall is the person?
___________m

Explanation / Answer

1) a )

Using the focal length formula

1/f = 1/d1 + 1/d 2

in the above equation d1 object distance = 27 - 2

= 25 cm(they are worn 2.0 cm from the eyes)

and in the above equation d2 is -48 - 2

= -46cm(they are worn 2.0 cm from the eyes)

The negative is because the image is virtual

1/f = 1/25 + 1/-46

1 /f = 0.04 - 0.021 = 0.019

f = 52.63cm

b )

in the above equation d1 object distance = 27 - 3

= 24 cm(they are worn 3.0 cm from the eyes)

and in the above equation d2 is -48 - 3

= -45cm(they are worn 3.0 cm from the eyes)

The negative is because the image is virtual

1/f = 1/24 + 1/-45

1 /f = 0.041 - 0.022 = 0.0191

f = 52.35cm

2) the converged lens will be project a real image of stamp at

1/11.4 = 1/si + 1/39.6

0.087 = 1/si + 0.025

1/si = 0.062

si = 16.12

for the magnification of the first image

m1 = - 39.6 / 16.12

= - 2.45

For the second lens, the image will be at

-1/6.06 = 1/si+ 1/ 33-16.12)

0.16= 1/si + 1/16.88
1/si = 0.16 - 0.059

1/si = 0.101

si = 9.9

The magnification of the image is

m2= - ( -9.9 / 16.88)

= 0.58
total magnification

m = m1m2

= 1.42

3)

1/s1 + 1/s2 = 1/f,

where s1 is the distance of the person behind the van from the lens,

s2 is the distance of the van owner eyes to the lens

1/s1 = -1/s2 + 1/f

1/s1 = 1/0.237 - 1/0.301

1/s1 = 4.219 - 3.32 = 0.899

s1 = 1.11 m

Person will be 1.11 meters behind the van of given

for height

1/H1 = 1/0.335 - 1/0.301

1/ H1 = 2.98 - 3.32

H1 = -0.34

= - 0.34, by means 0.34 m inverted image

so person 0.34 m tall

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