The Moon Through a Giant Telescope

solved ThinLens

Q. A giant refracting telescope at an observatory has an objective lens of focal length 15 m $f = 15$ m. The telescope is used to view the moon. The diameter of the moon is 3.48 × 10⁶ m $h_{o} = 3.48 \times 10^{6}$ m, and the radius of lunar orbit is 3.8 × 10⁸ m $d_{o} = 3.8 \times 10^{8}$ m. What is the diameter of the image of the moon formed by the objective lens, in cm, counted negative when inverted?

Inputs

focal length

f

15.00 m

moon

h_{o}

3480000.00 m

orbit

d_{o}

380000000.00 m

Outputs

image height

h_{i}

How it's solved

Solved with ThinLens

Given

$f = 15$ m
$h_{o} = 3.48 \times 10^{6}$ m
$d_{o} = 3.8 \times 10^{8}$ m

Find $h_{i}$ .

Working

an object in front is against the light(objective) $u = - d_{o}$
$u = - 3.8 \times 10^{8}$
$u = −3.8 \times 10^{8}$ m
thin lens formula(objective) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{(−3.8 \times 10^{8})} = \frac{1}{15}$
$v = 15.0000006$ m
magnification of a lens(objective) $m = \frac{v}{u}$
$m = \frac{15.0000006}{(−3.8 \times 10^{8})}$
$m = −3.94736858 \times 10^{−8}$
magnification compares the heights(objective) $m = \frac{h_{i}}{h_{o}}$
$(−3.94736858 \times 10^{−8}) = \frac{h_{i}}{3.48 \times 10^{6}}$
$h_{i} = −13.74$ cm
$h_{i} = −0.137368426$ m $= −13.7368426$ cm
$h_{i} \approx −13.7$ cmto 3 significant figures

Conclusion

The moon is so far away that its image forms at the focus of the objective, shrunk by the ratio of the focal length to the moon's distance and turned upside down.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :the_moon_through_a_giant_telescope do4  kind :numerical5  describes "A giant refracting telescope at an observatory has an objective lens of focal length " \6            "%{focal_length}. The telescope is used to view the moon. The diameter of the moon is %{moon}, and " \7            "the radius of lunar orbit is %{orbit}. %{image_height}, in cm, counted negative when inverted?"8  given focal_length: "15 m"9  given moon: "3.48e6 m"10  given orbit: "3.8e8 m"11  asks image_height: "What is the diameter of the image of the moon formed by the objective lens"12end

solution.rb

1require_relative "../questions/the_moon_through_a_giant_telescope"2require_relative "../science/ray_optics_and_optical_instruments/thin_lens"34solve :the_moon_through_a_giant_telescope, using: ThinLens do5  objects :objective67  take focal_length: objective.focal_length8  take moon: objective.object_height9  take orbit: objective.distance_in_front1011  image_height = objective.image_height12  answer image_height:, in: "cm", figures: 31314  conclusion "The moon is so far away that its image forms at the focus of the objective, shrunk by the ratio of " \15             "the focal length to the moon's distance and turned upside down."16end

playground.rb

1require_relative "../solutions/the_moon_through_a_giant_telescope"23play_with :the_moon_through_a_giant_telescope do4  title "The Moon Through a Giant Telescope"5end

The laws it drew on

quantity	written	units	within	from
refractive index of the lens	$n$	—	—	ThinLens
index of the surroundings	$n_{s}$	—	—	ThinLens
first radius	$R_{1}$	m	—	ThinLens
second radius	$R_{2}$	m	—	ThinLens
size of the second radius	$r_{2}$	m	—	ThinLens
focal length	$f$	m	—	ThinLens
size of the focal length	$f_{size}$	m	—	ThinLens
power of the lens	$P$	m⁻¹	—	ThinLens
distance in front	$d_{o}$	m	—	ThinLens
object distance	$u$	m	—	ThinLens
image distance	$v$	m	—	ThinLens
magnification	$m$	—	—	ThinLens
object height	$h_{o}$	m	—	ThinLens
image height	$h_{i}$	m	—	ThinLens
object distance of ground	$u_{g}$	m	—
distance in front of ground	$d_{o,g}$	m	—
object distance of objective	$u$	m	—
distance in front of objective	$d_{o}$	m	—
focal length of ground	$f_{g}$	m	—
refractive index of the lens of ground	$n_{g}$	—	—
index of the surroundings of ground	$n_{s,g}$	—	—
first radius of ground	$R_{1,g}$	m	—
second radius of ground	$R_{2,g}$	m	—
focal length of objective	$f$	m	—
refractive index of the lens of objective	$n$	—	—
index of the surroundings of objective	$n_{s}$	—	—
first radius of objective	$R_{1}$	m	—
second radius of objective	$R_{2}$	m	—
power of the lens of ground	$P_{g}$	m⁻¹	—
power of the lens of objective	$P$	m⁻¹	—
image distance of ground	$v_{g}$	m	—
image distance of objective	$v$	m	—
magnification of ground	$m_{g}$	—	—
magnification of objective	$m$	—	—
image height of ground	$h_{i,g}$	m	—
object height of ground	$h_{o,g}$	m	—
image height of objective	$h_{i}$	m	—
object height of objective	$h_{o}$	m	—
size of the second radius of objective	$r_{2}$	m	—
size of the focal length of objective	$f_{size}$	m	—

u_{l} = - d_{o,l}

an object in front is against the light

\frac{1}{f_{l}} = (\frac{n_{l}}{n_{s,l}} - 1) \cdot (\frac{1}{R_{1,l}} - \frac{1}{R_{2,l}})

lens makers formula

P_{l} = \frac{1}{f_{l}}

power of a lens

\frac{1}{v_{l}} - \frac{1}{u_{l}} = \frac{1}{f_{l}}

thin lens formula

m_{l} = \frac{v_{l}}{u_{l}}

magnification of a lens

m_{l} = \frac{h_{i,l}}{h_{o,l}}

magnification compares the heights

u_{g} = - d_{o,g}

an object in front is against the light(ground)

u = - d_{o}

an object in front is against the light(objective)

\frac{1}{f_{g}} = (\frac{n_{g}}{n_{s,g}} - 1) \cdot (\frac{1}{R_{1,g}} - \frac{1}{R_{2,g}})

lens makers formula(ground)

\frac{1}{f} = (\frac{n}{n_{s}} - 1) \cdot (\frac{1}{R_{1}} - \frac{1}{R_{2}})

lens makers formula(objective)

P_{g} = \frac{1}{f_{g}}

power of a lens(ground)

P = \frac{1}{f}

power of a lens(objective)

\frac{1}{v_{g}} - \frac{1}{u_{g}} = \frac{1}{f_{g}}

thin lens formula(ground)

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

thin lens formula(objective)

m_{g} = \frac{v_{g}}{u_{g}}

magnification of a lens(ground)

m = \frac{v}{u}

magnification of a lens(objective)

m_{g} = \frac{h_{i,g}}{h_{o,g}}

magnification compares the heights(ground)

m = \frac{h_{i}}{h_{o}}

magnification compares the heights(objective)