The Candle's Image

solved SphericalMirror

Q. A small candle, 2.5 cm $h_{o} = 2.5$ cm in size is placed at 27 cm $d_{o} = 27$ cm in front of a concave $R = - r_{size}$ mirror of radius of curvature 36 cm $r_{size} = 36$ cm. A screen is placed to obtain a sharp image. What is the size of the image, in cm, counted negative when inverted?

Inputs

size

h_{o}

2.5 cm

distance

d_{o}

27 cm

radius

r_{size}

36 cm

Outputs

image height

h_{i}

How it's solved

Solved with SphericalMirror

Given

$h_{o} = 2.5$ cm
$d_{o} = 27$ cm
$r_{size} = 36$ cm
$R = - r_{size}$ from “concave”

Find $h_{i}$ .

Working

$h_{o} = 2.5$ cm $= 0.025$ m
$d_{o} = 27$ cm $= 0.27$ m
$r_{size} = 36$ cm $= 0.36$ m
an object in front is against the light(mirror) $u = - d_{o}$
$u = - 0.27$
$u = −0.27$ m
concave $R = - r_{size}$
$R = - 0.36$
$R = −0.36$ m
focus halfway to the centre(mirror) $f = \frac{R}{2}$
$f = \frac{(−0.36)}{2}$
$f = −0.18$ m
mirror equation(mirror) $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{(−0.27)} = \frac{1}{(−0.18)}$
$v = −0.54$ m
magnification of a mirror(mirror) $m = \frac{- v}{u}$
$m = \frac{- (−0.54)}{(−0.27)}$
$m = −2$
magnification compares the heights(mirror) $m = \frac{h_{i}}{h_{o}}$
$(−2) = \frac{h_{i}}{0.025}$
$h_{i} = −5$ cm
$h_{i} = −0.05$ m $= −5$ cm
$h_{i} \approx −5.0$ cmto 2 significant figures

Conclusion

The image lies farther from the mirror than the candle does, so it is magnified; being real, it is inverted.

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_candle_s_image do4  kind :numerical5  describes "A small candle, %{size} in size is placed at %{distance} in front of a concave mirror of radius of " \6            "curvature %{radius}. A screen is placed to obtain a sharp image. %{image_height}, in cm, counted " \7            "negative when inverted?"8  given size: "2.5 cm"9  given distance: "27 cm"10  given radius: "36 cm"11  asks image_height: "What is the size of the image"12end

solution.rb

1require_relative "../questions/the_candle_s_image"2require_relative "../science/ray_optics_and_optical_instruments/spherical_mirror"34solve :the_candle_s_image, using: SphericalMirror do5  objects :mirror67  take size: mirror.object_height8  take distance: mirror.distance_in_front9  take radius: mirror.size_of_the_radius1011  deduce_from "concave", mirror.radius_of_curvature => -mirror.size_of_the_radius1213  image_height = mirror.image_height14  answer image_height:, in: "cm", figures: 21516  conclusion "The image lies farther from the mirror than the candle does, so it is magnified; being real, it is " \17             "inverted."18end

playground.rb

1require_relative "../solutions/the_candle_s_image"23play_with :the_candle_s_image do4  title "The Candle's Image"5end

The laws it drew on

quantity	written	units	within	from
radius of curvature	$R$	m	—	SphericalMirror
size of the radius	$r_{size}$	m	—	SphericalMirror
distance in front	$d_{o}$	m	—	SphericalMirror
focal length	$f$	m	—	SphericalMirror
size of the focal length	$f_{size}$	m	—	SphericalMirror
object distance	$u$	m	—	SphericalMirror
image distance	$v$	m	—	SphericalMirror
magnification	$m$	—	—	SphericalMirror
object height	$h_{o}$	m	—	SphericalMirror
image height	$h_{i}$	m	—	SphericalMirror
object distance of ground	$u_{g}$	m	—
distance in front of ground	$d_{o,g}$	m	—
object distance of mirror	$u$	m	—
distance in front of mirror	$d_{o}$	m	—
focal length of ground	$f_{g}$	m	—
radius of curvature of ground	$R_{g}$	m	—
focal length of mirror	$f$	m	—
radius of curvature of mirror	$R$	m	—
image distance of ground	$v_{g}$	m	—
image distance of mirror	$v$	m	—
magnification of ground	$m_{g}$	—	—
magnification of mirror	$m$	—	—
image height of ground	$h_{i,g}$	m	—
object height of ground	$h_{o,g}$	m	—
image height of mirror	$h_{i}$	m	—
object height of mirror	$h_{o}$	m	—
size of the radius of mirror	$r_{size}$	m	—
size of the focal length of mirror	$f_{size}$	m	—

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

an object in front is against the light

f_{m} = \frac{R_{m}}{2}

focus halfway to the centre

\frac{1}{v_{m}} + \frac{1}{u_{m}} = \frac{1}{f_{m}}

mirror equation

m_{m} = \frac{- v_{m}}{u_{m}}

magnification of a mirror

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

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(mirror)

f_{g} = \frac{R_{g}}{2}

focus halfway to the centre(ground)

f = \frac{R}{2}

focus halfway to the centre(mirror)

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

mirror equation(ground)

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

mirror equation(mirror)

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

magnification of a mirror(ground)

m = \frac{- v}{u}

magnification of a mirror(mirror)

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

magnification compares the heights(ground)

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

magnification compares the heights(mirror)

R = - r_{size}

concave