The Candle's Image

Q. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. A screen is placed to obtain a sharp image. What is the size of the image, in cm, counted negative when inverted? expects -5.0 cm

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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?

The playground

Inputs

size

h_{o}

2.5 cm

distance

d_{o}

27 cm

radius

r_{size}

36 cm

Outputs

image height

h_{i}

The solution

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 Ruby

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

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

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

The laws

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