The Needle's Magnification

Q. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. What is the magnification? expects 0.56

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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?

The playground

Inputs

size

h_{o}

4.5 cm

distance

d_{o}

12 cm

focal length

f_{size}

15 cm

Outputs

magnification

m

The solution

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 Ruby

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

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

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

f = f_{size}

convex mirror