The Image in a Concave Lens

solved ThinLens

Q. An object of size 3.0 cm $h_{o} = 3.0$ cm is placed 14 cm $d_{o} = 14$ cm in front of a concave lens $f = - f_{size}$ of focal length 21 cm $f_{size} = 21$ cm. What is the size of the image, in cm?

Inputs

size

h_{o}

3 cm

distance

d_{o}

14 cm

focal length

f_{size}

21 cm

Outputs

image height

h_{i}

How it's solved

Solved with ThinLens

Given

$h_{o} = 3.0$ cm
$d_{o} = 14$ cm
$f_{size} = 21$ cm
$f = - f_{size}$ from “concave lens”

Find $h_{i}$ .

Working

$h_{o} = 3.0$ cm $= 0.03$ m
$d_{o} = 14$ cm $= 0.14$ m
$f_{size} = 21$ cm $= 0.21$ m
an object in front is against the light(lens) $u = - d_{o}$
$u = - 0.14$
$u = −0.14$ m
concave lens $f = - f_{size}$
$f = - 0.21$
$f = −0.21$ m
thin lens formula(lens) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{(−0.14)} = \frac{1}{(−0.21)}$
$v = −0.084$ m
magnification of a lens(lens) $m = \frac{v}{u}$
$m = \frac{(−0.084)}{(−0.14)}$
$m = 0.6$
magnification compares the heights(lens) $m = \frac{h_{i}}{h_{o}}$
$0.6 = \frac{h_{i}}{0.03}$
$h_{i} = 1.8$ cm
$h_{i} = 0.018$ m $= 1.8$ cm
$h_{i} \approx 1.8$ cmto 2 significant figures

Conclusion

The image is nearer the lens than the object, so it is smaller; being virtual, it stands upright.

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_image_in_a_concave_lens do4  kind :numerical5  describes "An object of size %{size} is placed %{distance} in front of a concave lens of focal length " \6            "%{focal_length}. %{image_height}, in cm?"7  given size: "3.0 cm"8  given distance: "14 cm"9  given focal_length: "21 cm"10  asks image_height: "What is the size of the image"11end

solution.rb

1require_relative "../questions/the_image_in_a_concave_lens"2require_relative "../science/ray_optics_and_optical_instruments/thin_lens"34solve :the_image_in_a_concave_lens, using: ThinLens do5  objects :lens67  take size: lens.object_height8  take distance: lens.distance_in_front9  take focal_length: lens.size_of_the_focal_length1011  deduce_from "concave lens", lens.focal_length => -lens.size_of_the_focal_length1213  image_height = lens.image_height14  answer image_height:, in: "cm", figures: 21516  conclusion "The image is nearer the lens than the object, so it is smaller; being virtual, it stands upright."17end

playground.rb

1require_relative "../solutions/the_image_in_a_concave_lens"23play_with :the_image_in_a_concave_lens do4  title "The Image in a Concave Lens"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 lens	$u$	m	—
distance in front of lens	$d_{o}$	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 lens	$f$	m	—
refractive index of the lens of lens	$n$	—	—
index of the surroundings of lens	$n_{s}$	—	—
first radius of lens	$R_{1}$	m	—
second radius of lens	$R_{2}$	m	—
power of the lens of ground	$P_{g}$	m⁻¹	—
power of the lens of lens	$P$	m⁻¹	—
image distance of ground	$v_{g}$	m	—
image distance of lens	$v$	m	—
magnification of ground	$m_{g}$	—	—
magnification of lens	$m$	—	—
image height of ground	$h_{i,g}$	m	—
object height of ground	$h_{o,g}$	m	—
image height of lens	$h_{i}$	m	—
object height of lens	$h_{o}$	m	—
size of the second radius of lens	$r_{2}$	m	—
size of the focal length of lens	$f_{size}$	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(lens)

\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(lens)

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

power of a lens(ground)

P = \frac{1}{f}

power of a lens(lens)

\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(lens)

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

magnification of a lens(ground)

m = \frac{v}{u}

magnification of a lens(lens)

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

magnification compares the heights(ground)

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

magnification compares the heights(lens)

f = - f_{size}

concave lens