Question

The worldâs largest telescope made only from lenses (with no mirrors) is located at the Yerkes Observatory near Chicago. Its objective lens is 1.0 m in diameter and has a focal length of +18.9 m. The eyepiece has a focal length of +7.5 cm. The objective lens and eyepiece are separated by 18.970 m. (a) What is the location of the final image of a Moon crater 3.8Ã105km from Earth? (b) If the crater has a diameter of 2.0 km, what is the size of its final image? (c) Determine the angular magnification of the telescope by comparing the angular size of the image as seen through the telescope and the object as seen by the unaided eye.

Answer 1

a) Location of the final image = 1.05 m

b) Size of the final image = 0.00149 m

c) Angular Magnification = 270

Step-by-step explanation:

a )

Diameter of the objective lens = 1.0 m

Focal length of the objective lens = +18.9 m

the distance between the objective lens and the eyepiece = 18.970 m

The object forms a real image for the second lens at:

S₀ = 18.970 - 18.90

S₀ = 0.070 m

The focal length of the eyepiece, f = + 7.5 cm = 0.075 m

To calculate the location of the final image of the moon crater from the earth,
`S_(i)`. using the equation below:

`(1)/(f) = (1)/(S_(0) ) + \frac{1}{S{i} }`

`(1)/(0.075) - (1)/(0.070 ) + \frac{1}{S{i} }`

`(1)/(0.075) -(1)/(0.070 ) =\frac{1}{S{i} }`

`\frac{1}{S{i} } = -0.9524`

`S_(i) = -1.05 m`

taking the absolute value,
`S_(i) = 1.05 m`

b )

Diameter of the crater, D = 2.0 km = 2000 m

Distance of the crater from the earth, d = 3.8Ã10⁵km = 3.8 * 10⁵ * 10³

d = 3.8 * 10⁸ m

`S_(i) = 1.05 m`

focal length of the objective lens,
`f_(l) = +18.9 m`

Size of the final image =
`(D * S_(i)* f_(l) )/(dS_(0) )`

Size of the final image =
`(2000 * 1.05 * 18.9)/(3.8 * 10^(8) )`

Size of the final image =
`(39690)/(26600000)`

Size of the image = 0.00149 m

c )

Angular magnification of the telescope:

Magnification =
`(Final image size)/(Location of the final image) *(Distance for the moon crater to the earth)/(Diameter of the crater)`

Magnification =
`(0.00149)/(1.05) *(3.8 * 10^(8) )/(2000)`

Magnification = 270

Answer 2

Step-by-step explanation:

The objective lens is 1m

Focal length of the first lens

F1 = 18.9m

Focal length of the second lens

F2 = 7.5cm = 0.075m

Objective lens of the second lens is 18.97m

a. Location of final image?

The distance of the crater from the first lens is S1 = 3.8×10^5km

S1 = 3.8×10^8m.

S1' is the distance between the first image and the first lens

Using lens equation

1/S1 + 1/S1' = 1/F1

Then,

1/ 3.8×10^8 + 1/S1' = 1/18.9

1/S1' = 1/18.9 — 1/3.8×10^8

1/S1' = 0.0529

Then, S1' = 18.9m

Hence, the distance of the first image from the second lens is

S2 = 18.97—18.9

S2 = 0.07m

So, image produced by the first lens will behave as an object that is 0.07m from the second lens,

Also, let second lens final image be S2', then, applying lens equation we can determine S2'

1/S2 + 1/S2' = 1/F2

Then,

1/0.07 + 1/S2' = 1/0.075

1/S2' = 1/0.075 — 1/0.07

1/S2'= -0.9524

S2' = -1.05m

Since S2' if negative, this shows that the final image is located on the left side of the second lens

b. To determine the diameter of the final image, we need to find the linear magnification using magnification formulae

M = M1•M2

M = (S1'/S1)•(S2'/S2)

M = (18.9/3.8×10^8)•(-1.05/0.07)

M = 4.974×10^-8 × -15

M = —7.461 × 10^-7

Then, diameter of the final image is

d' = Md

Given that, the original diameter of the object d = 2km=2000m

d' = Md

d' = -7.461×10^-7 × 2000

d' = 1.4922 ×10^-3m

d' = 1.492mm

c. Angle of magnification and be determine by

M = θ'/θ

M = d'•S1/S2•d

M = -1.4922×10^-3 × 3.8×10^8/-1.05×2×10^3

M = 567,036/2100

M = 270