A Converging Beam Through a Convex Lens

Q. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is a convex lens of focal length 20 cm, in cm from the lens? expects 7.5 cm

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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?

The playground

Inputs

distance

u

12 cm

focal length

f_{size}

20 cm

Outputs

where

v

The solution

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 Ruby

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

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

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

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