The Needle's Magnification

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 magnification?

Inputs

size

h_{o}

4.5 cm

distance

d_{o}

12 cm

focal length

f_{size}

15 cm

Outputs

magnification

m

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 $m$ .

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) $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{(−0.12)} = \frac{1}{0.15}$
$v = 0.0666666667$ m
magnification of a mirror(mirror) $m = \frac{- v}{u}$
$m = \frac{- 0.0666666667}{(−0.12)}$
$m = 0.555555556$
$m \approx 0.56$ to 2 significant figures

Conclusion

A convex mirror always gives an upright image smaller than the object, so the magnification is positive and less than one.

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_needle_s_magnification do4  kind :numerical5  describes "A %{size} needle is placed %{distance} away from a convex mirror of focal length %{focal_length}. " \6            "%{magnification}?"7  given size: "4.5 cm"8  given distance: "12 cm"9  given focal_length: "15 cm"10  asks magnification: "What is the magnification"11end

solution.rb

1require_relative "../questions/the_needle_s_magnification"2require_relative "../science/ray_optics_and_optical_instruments/spherical_mirror"34solve :the_needle_s_magnification, 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  magnification = mirror.magnification14  answer magnification:, figures: 21516  conclusion "A convex mirror always gives an upright image smaller than the object, so the magnification is " \17             "positive and less than one."18end

playground.rb

1require_relative "../solutions/the_needle_s_magnification"23play_with :the_needle_s_magnification do4  title "The Needle's Magnification"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