The Image in a Concave Lens

Q. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. What is the size of the image, in cm? expects 1.8 cm

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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?

The playground

Inputs

size

h_{o}

3 cm

distance

d_{o}

14 cm

focal length

f_{size}

21 cm

Outputs

image height

h_{i}

The solution

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 Ruby

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

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

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

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