What is the diameter of the Andromeda galaxy (in light years)? If a person looks

Question

What is the diameter of the Andromeda galaxy (in light years)? If a person looks at a 30m tall oak tree that is 100 m away and the image of this tree appears on the cones & rods on the bank of their eye-ball 2.2 cm behind their eye lens. What is the focal distance of their eye? After enjoying looking at this tree she wants to read a book 20 cm away. To do this, her eyes will need to change focal distance by causing the lens to be... ...thicker in the middle. ...thinner in the middle. A microscope is made from two lenses. Each is convex with a focal distance of 4 cm. If only 1 lens is used and it is held 4.5 cm above a rock specimen, where should a paper be placed so that an image of the rock will appear on the paper? Now if the 1st lens is still 4.5 cm above the rock and the second lens is 40.2 cm above the first, Where should a paper be placed so an image of the rock magnified by both magnifying glasses be placed?

Explanation / Answer

Part 12)

for this problem we must use the equation builder

1/f = 1/o + 1/i

Part a)

where

o is the distance to the object

i is the image distance

f is the focal length

data

o = 100 m

i = 2.2 cm = 0.022 m

1/f = 1/ 100 + 1/0.022

1/f = (0.022 + 100) / 2.2

f = 2.2 / 100.022

f = 2.1995 10-2 m

f = 2.2 cm

the image is formed on the focal length

Part b)

o= 20 cm

usamos la misma ecuacion, calculamos i

1/i = 1/ f – 1/o

1/i = 1/ 2.2 - 1/ 20

1/i = (20-2.2)/44

i = 2.47 cm

we see that the image is formed later, so would remain unfocused, the eye for correguir this changes its focal length

1/f = 1/20 + 1/ 2.2

1/f = (2.2+20)/44

f = 1.98 cm

1/f = (n-1) ( 1/R)

As we see the foal distance decreases expression and the focal length is inversely proportional to the radius, whereby the lens is thicker in the middle part

result i

Part 13)

part a

as we have a single lens is used, we use the same expression builder

1/i = 1/f -1/o

1/i = 1/ 4 – 1/ 4.5

1/i = (4.5 -4) / 18

i = 36 cm

part b)

In this case the lens system behaves like a microscope, we calculate the magnification with the expression

M = Mome= -L/fo (25 cm/fe)

data

L =40.2 cm

fo =4 cm

fe =4 cm

The value of 25 cm is the point of near vision of a normal person

M = - 40.2/ 4 (25/4)

M = 62.5

We calculate the position of the image, we look for this image of the first lens and use this image as an object for the second lens, remember that distances are measured relative lens analysis

for the first lens i = 36 cm

data for the second lens

o2 =40.2 -36 = 4.2 cm

1/i = 1/f – 1/o2

1/i =1/ 4 - 1/ 4.2

1/i = (4.2 -4) /16.8

i = 84 cm

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.