A Converging Beam Through a Convex Lens

solved ThinLens

Q. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm $u = 12$ cm from P. At what point does the beam converge if the lens is a convex lens $f = f_{size}$ of focal length 20 cm $f_{size} = 20$ cm, in cm from the lens?

Inputs

distance

u

12 cm

focal length

f_{size}

20 cm

Outputs

where

v

How it's solved

Solved with ThinLens

Given

$u = 12$ cm
$f_{size} = 20$ cm
$f = f_{size}$ from “convex lens”

Find $v$ .

Working

$u = 12$ cm $= 0.12$ m
$f_{size} = 20$ cm $= 0.2$ m
convex lens $f = f_{size}$
$f = 0.2$
$f = 0.2$ m
thin lens formula(lens) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{0.12} = \frac{1}{0.2}$
$v = 7.5$ cm
$v = 0.075$ m $= 7.5$ cm
$v \approx 7.5$ cmto 2 significant figures

Conclusion

The point the beam was heading for lies beyond the lens, so it is a virtual object at a positive distance; a convex lens pulls the beam to a point nearer still.

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_converging_beam_through_a_convex_lens do4  kind :numerical5  describes "A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam " \6            "%{distance} from P. %{where} if the lens is a convex lens of focal length %{focal_length}, in cm " \7            "from the lens?"8  given distance: "12 cm"9  given focal_length: "20 cm"10  asks where: "At what point does the beam converge"11end

solution.rb

1require_relative "../questions/a_converging_beam_through_a_convex_lens"2require_relative "../science/ray_optics_and_optical_instruments/thin_lens"34solve :a_converging_beam_through_a_convex_lens, using: ThinLens do5  objects :lens67  take distance: lens.object_distance8  take focal_length: lens.size_of_the_focal_length910  deduce_from "convex lens", lens.focal_length => lens.size_of_the_focal_length1112  where = lens.image_distance13  answer where:, in: "cm", figures: 21415  conclusion "The point the beam was heading for lies beyond the lens, so it is a virtual object at a positive " \16             "distance; a convex lens pulls the beam to a point nearer still."17end

playground.rb

1require_relative "../solutions/a_converging_beam_through_a_convex_lens"23play_with :a_converging_beam_through_a_convex_lens do4  title "A Converging Beam Through a Convex Lens"5end

The laws it drew on

quantity	written	units	within	from
refractive index of the lens	$n$	—	—	ThinLens
index of the surroundings	$n_{s}$	—	—	ThinLens
first radius	$R_{1}$	m	—	ThinLens
second radius	$R_{2}$	m	—	ThinLens
size of the second radius	$r_{2}$	m	—	ThinLens
focal length	$f$	m	—	ThinLens
size of the focal length	$f_{size}$	m	—	ThinLens
power of the lens	$P$	m⁻¹	—	ThinLens
distance in front	$d_{o}$	m	—	ThinLens
object distance	$u$	m	—	ThinLens
image distance	$v$	m	—	ThinLens
magnification	$m$	—	—	ThinLens
object height	$h_{o}$	m	—	ThinLens
image height	$h_{i}$	m	—	ThinLens
object distance of ground	$u_{g}$	m	—
distance in front of ground	$d_{o,g}$	m	—
object distance of lens	$u$	m	—
distance in front of lens	$d_{o}$	m	—
focal length of ground	$f_{g}$	m	—
refractive index of the lens of ground	$n_{g}$	—	—
index of the surroundings of ground	$n_{s,g}$	—	—
first radius of ground	$R_{1,g}$	m	—
second radius of ground	$R_{2,g}$	m	—
focal length of lens	$f$	m	—
refractive index of the lens of lens	$n$	—	—
index of the surroundings of lens	$n_{s}$	—	—
first radius of lens	$R_{1}$	m	—
second radius of lens	$R_{2}$	m	—
power of the lens of ground	$P_{g}$	m⁻¹	—
power of the lens of lens	$P$	m⁻¹	—
image distance of ground	$v_{g}$	m	—
image distance of lens	$v$	m	—
magnification of ground	$m_{g}$	—	—
magnification of lens	$m$	—	—
image height of ground	$h_{i,g}$	m	—
object height of ground	$h_{o,g}$	m	—
image height of lens	$h_{i}$	m	—
object height of lens	$h_{o}$	m	—
size of the second radius of lens	$r_{2}$	m	—
size of the focal length of lens	$f_{size}$	m	—

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

an object in front is against the light

\frac{1}{f_{l}} = (\frac{n_{l}}{n_{s,l}} - 1) \cdot (\frac{1}{R_{1,l}} - \frac{1}{R_{2,l}})

lens makers formula

P_{l} = \frac{1}{f_{l}}

power of a lens

\frac{1}{v_{l}} - \frac{1}{u_{l}} = \frac{1}{f_{l}}

thin lens formula

m_{l} = \frac{v_{l}}{u_{l}}

magnification of a lens

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

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(lens)

\frac{1}{f_{g}} = (\frac{n_{g}}{n_{s,g}} - 1) \cdot (\frac{1}{R_{1,g}} - \frac{1}{R_{2,g}})

lens makers formula(ground)

\frac{1}{f} = (\frac{n}{n_{s}} - 1) \cdot (\frac{1}{R_{1}} - \frac{1}{R_{2}})

lens makers formula(lens)

P_{g} = \frac{1}{f_{g}}

power of a lens(ground)

P = \frac{1}{f}

power of a lens(lens)

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

thin lens formula(ground)

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

thin lens formula(lens)

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

magnification of a lens(ground)

m = \frac{v}{u}

magnification of a lens(lens)

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

magnification compares the heights(ground)

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

magnification compares the heights(lens)

f = f_{size}

convex lens