A Needle Before a Convex Mirror

solved SphericalMirror

Q. A 4.5 cm $h_{o} = 4.5$ cm needle is placed 12 cm $d_{o} = 12$ cm away from a convex mirror $f = f_{size}$ of focal length 15 cm $f_{size} = 15$ cm. What is the location of the image, in cm, measured along the incoming light from the mirror?

Inputs

size

h_{o}

4.5 cm

distance

d_{o}

12 cm

focal length

f_{size}

15 cm

Outputs

where

v

How it's solved

Solved with SphericalMirror

Given

$h_{o} = 4.5$ cm
$d_{o} = 12$ cm
$f_{size} = 15$ cm
$f = f_{size}$ from “convex mirror”

Find $v$ .

Working

$h_{o} = 4.5$ cm $= 0.045$ m
$d_{o} = 12$ cm $= 0.12$ m
$f_{size} = 15$ cm $= 0.15$ m
an object in front is against the light(mirror) $u = - d_{o}$
$u = - 0.12$
$u = −0.12$ m
convex mirror $f = f_{size}$
$f = 0.15$
$f = 0.15$ m
mirror equation(mirror) $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{(−0.12)} = \frac{1}{0.15}$
$v = 6.667$ cm
$v = 0.0666666667$ m $= 6.66666667$ cm
$v \approx 6.7$ cmto 2 significant figures

Conclusion

A convex mirror's focus lies behind it, so the image is virtual, behind the mirror and between the pole and the focus.

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_needle_before_a_convex_mirror do4  kind :numerical5  describes "A %{size} needle is placed %{distance} away from a convex mirror of focal length %{focal_length}. " \6            "%{where}, in cm, measured along the incoming light from the mirror?"7  given size: "4.5 cm"8  given distance: "12 cm"9  given focal_length: "15 cm"10  asks where: "What is the location of the image"11end

solution.rb

1require_relative "../questions/a_needle_before_a_convex_mirror"2require_relative "../science/ray_optics_and_optical_instruments/spherical_mirror"34solve :a_needle_before_a_convex_mirror, using: SphericalMirror do5  objects :mirror67  take size: mirror.object_height8  take distance: mirror.distance_in_front9  take focal_length: mirror.size_of_the_focal_length1011  deduce_from "convex mirror", mirror.focal_length => mirror.size_of_the_focal_length1213  where = mirror.image_distance14  answer where:, in: "cm", figures: 21516  conclusion "A convex mirror's focus lies behind it, so the image is virtual, behind the mirror and between the " \17             "pole and the focus."18end

playground.rb

1require_relative "../solutions/a_needle_before_a_convex_mirror"23play_with :a_needle_before_a_convex_mirror do4  title "A Needle Before a Convex Mirror"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)

f = f_{size}

convex mirror