A Tower Through a Small Telescope

solved ThinLens

Q. A small telescope has an objective lens of focal length 140 cm $f = 140$ cm. The telescope is used to view a 100 m $h_{o} = 100$ m tall tower 3 km $d_{o} = 3$ km away. What is the height of the image of the tower formed by the objective lens, in cm, counted negative when inverted?

Inputs

focal length

f

140 cm

height

h_{o}

100.00 m

distance

d_{o}

3 km

Outputs

image height

h_{i}

How it's solved

Solved with ThinLens

Given

$f = 140$ cm
$h_{o} = 100$ m
$d_{o} = 3$ km

Find $h_{i}$ .

Working

$f = 140$ cm $= 1.4$ m
$d_{o} = 3$ km $= 3000$ m
an object in front is against the light(objective) $u = - d_{o}$
$u = - 3000$
$u = −3000$ m
thin lens formula(objective) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{(−3000)} = \frac{1}{1.4}$
$v = 1.40065364$ m
magnification of a lens(objective) $m = \frac{v}{u}$
$m = \frac{1.40065364}{(−3000)}$
$m = −4.66884546 \times 10^{−4}$
magnification compares the heights(objective) $m = \frac{h_{i}}{h_{o}}$
$(−4.66884546 \times 10^{−4}) = \frac{h_{i}}{100}$
$h_{i} = −4.669$ cm
$h_{i} = −0.0466884546$ m $= −4.66884546$ cm
$h_{i} \approx −4.7$ cmto 2 significant figures

Conclusion

The tower is far compared with the focal length, so its real, inverted image forms almost at the focus, scaled down by the ratio of the two distances.

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 :a_tower_through_a_small_telescope do4  kind :numerical5  describes "A small telescope has an objective lens of focal length %{focal_length}. The telescope is used to " \6            "view a %{height} tall tower %{distance} away. %{image_height}, in cm, counted negative when " \7            "inverted?"8  given focal_length: "140 cm"9  given height: "100 m"10  given distance: "3 km"11  asks image_height: "What is the height of the image of the tower formed by the objective lens"12end

solution.rb

1require_relative "../questions/a_tower_through_a_small_telescope"2require_relative "../science/ray_optics_and_optical_instruments/thin_lens"34solve :a_tower_through_a_small_telescope, using: ThinLens do5  objects :objective67  take focal_length: objective.focal_length8  take height: objective.object_height9  take distance: objective.distance_in_front1011  image_height = objective.image_height12  answer image_height:, in: "cm", figures: 21314  conclusion "The tower is far compared with the focal length, so its real, inverted image forms almost at the " \15             "focus, scaled down by the ratio of the two distances."16end

playground.rb

1require_relative "../solutions/a_tower_through_a_small_telescope"23play_with :a_tower_through_a_small_telescope do4  title "A Tower Through a Small 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)