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GRADE 11 PRACTICE

Physics Problems: Optics, Grade 11

Reflection, refraction, lenses and diffraction.

24 problems across 4 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

24 problems
Position in this set: 1 from 24

Problem #37

Geometric optics

Beam angle 40^{\circ}, refractive index of the medium 1{,}5. What is the angle of refraction? Light falls from the air; take its refractive index equal to 1.

Answer options

  1. 50{,}8\,^{\circ}
  2. 254\,^{\circ}
  3. 25{,}4\,^{\circ}
  4. 12{,}7\,^{\circ}
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Given
\alpha
40^{\circ}
n
1{,}5
Find

\beta

  1. n = \dfrac{\sin\alpha}{\sin\beta}\ \Rightarrow\ \sin\beta = \dfrac{\sin\alpha}{n}\dfrac{\sin 40^{\circ}}{1{,}5} = \dfrac{0{,}643}{1{,}5}\sin\beta = 0{,}4285
  2. \beta = \arcsin 0{,}4285\beta = 25{,}4^{\circ}

Answer\beta = 25{,}4^{\circ}

Position in this set: 2 from 24

Problem #38

Lenses, interference and diffraction

Distance from an object to an image of equal size 80\,\text{см}. Determine the focal length and optical power of the lens.

Answer options

  1. \left(0{,}1\,\text{м};\ 2{,}5\,\text{дптр}\right)
  2. \left(0{,}4\,\text{м};\ 10\,\text{дптр}\right)
  3. \left(2\,\text{м};\ 50\,\text{дптр}\right)
  4. \left(0{,}2\,\text{м};\ 5\,\text{дптр}\right)
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Given
S
80\,\text{см}
\Gamma
1
Find

FD

  1. \Gamma = \dfrac{f}{d} = 1\ \Rightarrow\ f = d

    The image is equal to the object only at d = 2F.

  2. \dfrac{1}{F} = \dfrac{1}{d} + \dfrac{1}{f} = \dfrac{2}{d}\ \Rightarrow\ d = 2F,\ f = 2F
  3. S = d + f = 4F\ \Rightarrow\ F = \dfrac{S}{4}F = \dfrac{80}{4}F = 20\,\text{см} = 0{,}2\,\text{м}
  4. D = \dfrac{1}{F}D = \dfrac{1}{0{,}2}D = 5\,\text{дптр}

AnswerF = 0{,}2\,\text{м}D = 5\,\text{дптр}

Position in this set: 3 from 24

Problem #41

Geometric optics

Height of the Sun above the horizon 20^{\circ}. At what angle to the horizon should a mirror be placed to illuminate the bottom of a vertical well?

Answer options

  1. 110\,^{\circ}
  2. 550\,^{\circ}
  3. 55\,^{\circ}
  4. 27{,}5\,^{\circ}
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Given
h_{\odot}
20^{\circ}
\alpha
70^{\circ}
Find

\beta

  1. \alpha_1 = \gamma_1 = \dfrac{90^{\circ} + 20^{\circ}}{2}\beta = 55^{\circ}

    The beam must go vertically downwards, so the angle between the incident and reflected rays is 110°.

  2. \beta = \gamma\beta = 55^{\circ}

    Angles with mutually perpendicular sides are equal: the mirror is placed at 55° to the horizon.

Answer\beta = 55^{\circ}

Position in this set: 4 from 24

Problem #42

Geometric optics

At angle of incidence 40^{\circ} the angle of refraction is 24^{\circ}. Determine the refractive index, the speed of light in the substance, and the limiting angle of total internal reflection. What will happen at the angle of incidence 40^{\circ} from matter into air?

Answer options

  1. \left(n=1{,}5;\ 2\cdot10^8\,\text{м/с};\ 41{,}8^{\circ};\ \text{преломление}\right)
  2. \left(n=0{,}63;\ 4{,}8\cdot10^8\,\text{м/с};\ 50^{\circ};\ \text{отражения нет}\right)
  3. \left(n=1{,}58;\ 3\cdot10^8\,\text{м/с};\ 90^{\circ};\ \text{преломление}\right)
  4. \left(n=1{,}58;\ 1{,}9\cdot10^8\,\text{м/с};\ 39{,}3^{\circ};\ \text{полное отражение}\right)
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Given
\alpha
40^{\circ}
\beta
24^{\circ}
\beta_2
40^{\circ}
c
3\cdot 10^{8}\,\text{м/с}
Find

nc_{\text{в}}\alpha_0\alpha_2

  1. n = \dfrac{\sin\alpha}{\sin\beta}\dfrac{\sin 40^{\circ}}{\sin 24^{\circ}} = \dfrac{0{,}6428}{0{,}4067}n = 1{,}58
  2. c_{\text{в}} = \dfrac{c}{n}c_{\text{в}} = \dfrac{3\cdot 10^{8}}{1{,}58}c_{\text{в}} = 1{,}9\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}
  3. \sin\alpha_0 = \dfrac{1}{n}\sin\alpha_0 = \dfrac{1}{1{,}58}\alpha_0 = 0{,}6328\ \Rightarrow\ \alpha_0 = 39{,}3^{\circ}

    Limiting angle of total reflection.

  4. \sin\alpha_2 = n\sin\beta_2\sin\alpha_2 = 1{,}58\cdot 0{,}64281{,}0156 > 1

    There is no angle of incidence with 40° refraction: the maximum angle of refraction is 39.3°, and this is achieved at 90° incidence.

Answern = 1{,}58c_{\text{в}} = 1{,}9\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}\alpha_0 = 39{,}3^{\circ}\text{When fall from substances B air at an angle 40° is happening Total internal reflection}

Position in this set: 5 from 24

Problem #43

Geometric optics

What must be the refractive index of a substance so that the reflected and refracted rays make an angle? 90^{\circ}? Express your answer in terms of the angle of incidence \alpha. Light falls from the air; take the refractive index of air equal to 1.

Answer options

  1. n=\tan\alpha,\ n\text{ безразмерен}
  2. n=\sin\alpha,\ n\text{ безразмерен}
  3. n=\cos\alpha,\ n\text{ безразмерен}
  4. n=\cot\alpha,\ n\text{ безразмерен}
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Given
\beta + \gamma
90^{\circ}
Find

n

  1. n = \dfrac{\sin\alpha}{\sin\beta} = \dfrac{\sin\alpha}{\sin\left(90^{\circ} - \alpha\right)}

    The angle of reflection is equal to the angle of incidence, therefore, when the rays are perpendicular, β = 90° − α.

  2. n = \dfrac{\sin\alpha}{\cos\alpha} = \operatorname{tg}\alphan = \operatorname{tg}\alpha

    For example, at α = 60° you get n = tan 60° ≈ 1.73.

Answern = \operatorname{tg}\alpha

Position in this set: 6 from 24

Problem #44

Lenses, interference and diffraction

The object is at a distance 20\,\text{см} from the lens, the image in 3 times the size of the item. Determine the possible focal lengths and optical powers for real and virtual images.

Answer options

  1. \left(150\,\text{см};\ 66{,}7\,\text{дптр};\ 300\,\text{см};\ 33{,}3\,\text{дптр}\right)
  2. \left(15\,\text{см};\ 6{,}67\,\text{дптр};\ 30\,\text{см};\ 3{,}33\,\text{дптр}\right)
  3. \left(7{,}5\,\text{см};\ 3{,}335\,\text{дптр};\ 15\,\text{см};\ 1{,}665\,\text{дптр}\right)
  4. \left(30\,\text{см};\ 13{,}34\,\text{дптр};\ 60\,\text{см};\ 6{,}66\,\text{дптр}\right)
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Given
d
20\,\text{см}
\Gamma
3
Find

F_1D_1F_2D_2

  1. \Gamma = \dfrac{\left|f\right|}{d}\ \Rightarrow\ \left|f\right| = \Gamma d\left|f\right| = 3\cdot 20\left|f\right| = 60\,\text{см}
  2. \dfrac{1}{F} = \dfrac{1}{d} + \dfrac{1}{f}\ \Rightarrow\ F = \dfrac{df}{d + f}
  3. F_1 = \dfrac{20\cdot 60}{20 + 60}F_1 = 15\,\text{см} = 0{,}15\,\text{м}

    The image is valid.

  4. D_1 = \dfrac{1}{F_1}D_1 = \dfrac{1}{0{,}15}D_1 = 6{,}67\,\text{дптр}
  5. F_2 = \dfrac{20\cdot\left(-60\right)}{20 + \left(-60\right)}F_2 = 30\,\text{см} = 0{,}3\,\text{м}

    The image is imaginary - for it f is taken with a minus sign.

  6. D_2 = \dfrac{1}{F_2}D_2 = \dfrac{1}{0{,}3}D_2 = 3{,}33\,\text{дптр}

AnswerF_1 = 15\,\text{см}D_1 = 6{,}67\,\text{дптр}F_2 = 30\,\text{см}D_2 = 3{,}33\,\text{дптр}

Position in this set: 7 from 24

Problem #45

Lenses, interference and diffraction

Converging lens with focal length 20\,\text{см} gives an image enlarged by 5 times. Find the distance between the object and the image: a) for a real image; b) for imaginary. Then find this distance if the actual image is reduced in 5 times.

Answer options

  1. \left(288\,\text{см};\ 128\,\text{см};\ 288\,\text{см}\right)
  2. \left(1440\,\text{см};\ 640\,\text{см};\ 1440\,\text{см}\right)
  3. \left(144\,\text{см};\ 64\,\text{см};\ 144\,\text{см}\right)
  4. \left(72\,\text{см};\ 32\,\text{см};\ 72\,\text{см}\right)
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Given
F
20\,\text{см}
\Gamma
5
\Gamma_2
\dfrac{1}{5}
Find

S_1S_2S_3

  1. \dfrac{1}{F} = \dfrac{1}{d} + \dfrac{1}{\Gamma d} = \dfrac{\Gamma + 1}{\Gamma d}\ \Rightarrow\ d_1 = \dfrac{F\left(\Gamma + 1\right)}{\Gamma}d_1 = \dfrac{20\left(5 + 1\right)}{5}d_1 = 24\,\text{см}
  2. f_1 = \Gamma d_1,\quad S_1 = d_1 + f_1f_1 = 120\,\text{см};\quad S_1 = 24 + 120S_1 = 144\,\text{см}

    Real image: object and image on opposite sides of the lens.

  3. \dfrac{1}{F} = \dfrac{1}{d} - \dfrac{1}{\Gamma d} = \dfrac{\Gamma - 1}{\Gamma d}\ \Rightarrow\ d_2 = \dfrac{F\left(\Gamma - 1\right)}{\Gamma}d_2 = \dfrac{20\left(5 - 1\right)}{5}d_2 = 16\,\text{см}
  4. S_2 = \left|f_2\right| - d_2S_2 = 80 - 16S_2 = 64\,\text{см}

    Virtual image: it is on the same side as the object.

  5. \Gamma_2 = \dfrac{1}{5}:\quad d_3 = \dfrac{F\left(\Gamma_2 + 1\right)}{\Gamma_2}\dfrac{20\cdot 1{,}2}{0{,}2} = 120\,\text{см};\quad f_3 = 24\,\text{см}S_3 = 144\,\text{см}

    Zooming out by 5 times is the same scheme as zooming in, only the object and image are swapped.

AnswerS_1 = 144\,\text{см}S_2 = 64\,\text{см}S_3 = 144\,\text{см}

Position in this set: 8 from 24

Problem #46

Lenses, interference and diffraction

Coherent waves with a frequency interfere in the air 5\cdot10^{14}\,\text{Гц}. The light will increase or decrease at a point with a path difference 2{,}4\,\text{мкм}? The sources are in phase, the wave polarizations are the same.

Answer options

  1. \Delta r=7\dfrac{\lambda}{2},\ \text{ослабление}
  2. \Delta r=4\lambda,\ \text{ослабление}
  3. \Delta r=3\dfrac{\lambda}{2},\ \text{усиление}
  4. \Delta r=8\dfrac{\lambda}{2},\ \text{усиление}
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Given
\nu
5\cdot 10^{14}\,\text{Гц}
\Delta d
2{,}4\cdot 10^{-6}\,\text{м}
c
3\cdot 10^{8}\,\text{м/с}
Find

\dfrac{\Delta d}{\lambda/2}

  1. \lambda = \dfrac{c}{\nu}\lambda = \dfrac{3\cdot 10^{8}}{5\cdot 10^{14}}6\cdot 10^{-7}\,\text{м}\ \Rightarrow\ \dfrac{\lambda}{2} = 3\cdot 10^{-7}\,\text{м}
  2. \dfrac{\Delta d}{\lambda/2}\dfrac{\Delta d}{\lambda/2} = \dfrac{2{,}4\cdot 10^{-6}}{3\cdot 10^{-7}}\dfrac{\Delta d}{\lambda/2} = 8

    An even number of half-waves means the light will intensify: this is the maximum.

Answer\dfrac{\Delta d}{\lambda/2} = 8\ \text{- свет усилится}

Position in this set: 9 from 24

Problem #47

Lenses, interference and diffraction

On a diffraction grating with a period 1/200\,\text{мм} Monochromatic light falls normally. The third order maximum is visible at an angle 17{,}5^{\circ}. Find the wavelength in air and the wavelength and speed of light in glass with n=1{,}5, as well as the original frequency. Then find the wavelengths in air and glass at twice the frequency, assuming the refractive index remains constant.

Answer options

  1. \left(5 \cdot 10^{-7}\,\text{м};\ 3{,}33 \cdot 10^{-7}\,\text{м};\ 2 \cdot 10^{8}\,\text{м/с};\ 6 \cdot 10^{14}\,\text{Гц};\ 2{,}5 \cdot 10^{-7}\,\text{м};\ 1{,}67 \cdot 10^{-7}\,\text{м}\right)
  2. \left(2{,}5 \cdot 10^{-7}\,\text{м};\ 1{,}665 \cdot 10^{-7}\,\text{м};\ 1 \cdot 10^{8}\,\text{м/с};\ 3 \cdot 10^{14}\,\text{Гц};\ 1{,}25 \cdot 10^{-7}\,\text{м};\ 8{,}35 \cdot 10^{-8}\,\text{м}\right)
  3. \left(1 \cdot 10^{-6}\,\text{м};\ 6{,}66 \cdot 10^{-7}\,\text{м};\ 4 \cdot 10^{8}\,\text{м/с};\ 1{,}2 \cdot 10^{15}\,\text{Гц};\ 5 \cdot 10^{-7}\,\text{м};\ 3{,}34 \cdot 10^{-7}\,\text{м}\right)
  4. \left(5 \cdot 10^{-6}\,\text{м};\ 3{,}33 \cdot 10^{-6}\,\text{м};\ 2 \cdot 10^{9}\,\text{м/с};\ 6 \cdot 10^{15}\,\text{Гц};\ 2{,}5 \cdot 10^{-6}\,\text{м};\ 1{,}67 \cdot 10^{-6}\,\text{м}\right)
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Given
d
\dfrac{1}{200}\,\text{мм}0{,}005\cdot 10^{-3}\,\text{м}
k
3
\alpha
17{,}5^{\circ}
n
1{,}5
\nu_1
2\nu
Find

\lambda\lambda_{\text{ст}}c_{\text{ст}}\nu\lambda_1\lambda_{1\text{ст}}

  1. d\sin\alpha = k\lambda\ \Rightarrow\ \lambda = \dfrac{d\sin\alpha}{k}\lambda = \dfrac{0{,}005\cdot 10^{-3}\cdot\sin 17{,}5^{\circ}}{3}\lambda = 5\cdot 10^{-7}\,\text{м}
  2. \lambda_{\text{ст}} = \dfrac{\lambda}{n}\lambda_{\text{ст}} = \dfrac{5\cdot 10^{-7}}{1{,}5}\lambda_{\text{ст}} = 3{,}33\cdot 10^{-7}\,\text{м}
  3. c_{\text{ст}} = \dfrac{c}{n}c_{\text{ст}} = \dfrac{3\cdot 10^{8}}{1{,}5}c_{\text{ст}} = 2\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}
  4. \nu = \dfrac{c}{\lambda}\nu = \dfrac{3\cdot 10^{8}}{5\cdot 10^{-7}}\nu = 6\cdot 10^{14}\,\text{Гц}

    The frequency does not change when passing into glass - the speed and wavelength change.

  5. \lambda_1 = \dfrac{c}{\nu_1} = \dfrac{c}{2\nu} = \dfrac{\lambda}{2}\lambda_1 = \dfrac{5\cdot 10^{-7}}{2}\lambda_1 = 2{,}5\cdot 10^{-7}\,\text{м}

    Doubling the frequency shortens the wave by half - this is ultraviolet.

  6. \lambda_{1\text{ст}} = \dfrac{\lambda_1}{n}\lambda_{1\text{ст}} = \dfrac{2{,}5\cdot 10^{-7}}{1{,}5}\lambda_{1\text{ст}} = 1{,}67\cdot 10^{-7}\,\text{м}

Answer\lambda = 5\cdot 10^{-7}\,\text{м}\lambda_{\text{ст}} = 3{,}33\cdot 10^{-7}\,\text{м}c_{\text{ст}} = 2\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}\nu = 6\cdot 10^{14}\,\text{Гц}\lambda_1 = 2{,}5\cdot 10^{-7}\,\text{м}\lambda_{1\text{ст}} = 1{,}67\cdot 10^{-7}\,\text{м}

Position in this set: 10 from 24

Problem #48

Lenses, interference and diffraction

The diffraction grating has 85 cracks on 1\,\text{мм}. What is the angle between the second order purple maxima at \lambda=4\cdot10^{-7}\,\text{м}?

Answer options

  1. 78\,^{\circ}
  2. 7{,}8\,^{\circ}
  3. 3{,}9\,^{\circ}
  4. 15{,}6\,^{\circ}
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Given
d
\dfrac{1}{85}\,\text{мм}1{,}176\cdot 10^{-5}\,\text{м}
k
2
\lambda
4\cdot 10^{-7}\,\text{м}
Find

\gamma

  1. d\sin\alpha = k\lambda\ \Rightarrow\ \sin\alpha = \dfrac{k\lambda}{d}\sin\alpha = \dfrac{2\cdot 4\cdot 10^{-7}}{1{,}176\cdot 10^{-5}}\sin\alpha = 0{,}068
  2. \alpha = \arcsin 0{,}068\alpha = 3{,}9^{\circ}
  3. \gamma = 2\alpha\gamma = 2\cdot 3{,}9\gamma = 7{,}8^{\circ}

    The maxima are symmetrical about the center, so the angle between them is twice as large.

Answer\gamma = 7{,}8^{\circ}

Position in this set: 11 from 24

Problem #49

Lenses, interference and diffraction

Focal length of the projection lamp lens 15\,\text{см}. The slide is at a distance 15{,}6\,\text{см} from the lens. What linear magnification does the flashlight provide?

Answer options

  1. 50\,\text{раз}
  2. 250\,\text{раз}
  3. 25\,\text{раз}
  4. 12{,}5\,\text{раз}
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Given
F
15\,\text{см}
d
15{,}6\,\text{см}
Find

\Gamma

  1. \dfrac{1}{F} = \dfrac{1}{d} + \dfrac{1}{f}\ \Rightarrow\ f = \dfrac{dF}{d - F}f = \dfrac{15{,}6\cdot 15}{15{,}6 - 15}f = 390\,\text{см}
  2. \Gamma = \dfrac{f}{d}\Gamma = \dfrac{390}{15{,}6}\Gamma = 25

    The slide is placed a little further from the focus - then the image turns out large and real.

Answer\Gamma = 25

Position in this set: 12 from 24

Problem #52

Lenses, interference and diffraction

At a distance 1{,}5\,\text{м} from the screen there is a diffraction grating with a period 20\,\text{мкм}. At what distance from the central maximum is the second-order violet maximum at \lambda=4\cdot10^{-7}\,\text{м}?

Answer options

  1. 0{,}6\,\text{м}
  2. 0{,}06\,\text{м}
  3. 0{,}03\,\text{м}
  4. 0{,}12\,\text{м}
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Given
L
1{,}5\,\text{м}
d
20\,\text{мкм}20\cdot 10^{-6}\,\text{м}
k
2
\lambda
4\cdot 10^{-7}\,\text{м}
Find

\ell

  1. d\sin\alpha = k\lambda,\quad \sin\alpha \approx \operatorname{tg}\alpha = \dfrac{\ell}{L}

    The angles are small, so the sine can be replaced by a tangent.

  2. \dfrac{d\ell}{L} = k\lambda\ \Rightarrow\ \ell = \dfrac{k\lambda L}{d}\ell = \dfrac{2\cdot 4\cdot 10^{-7}\cdot 1{,}5}{20\cdot 10^{-6}}\ell = 0{,}06\,\text{м} = 6\,\text{см}

Answer\ell = 0{,}06\,\text{м}

Position in this set: 13 from 24

Problem #122

Reflection and refraction of light

Refractive index of the medium 1{,}5. Speed of light in vacuum 300000000 m/s. Find the speed of light in the medium.

Answer options

  1. 200000000\,\text{м/с}
  2. 400000000\,\text{м/с}
  3. 100000000\,\text{м/с}
  4. 600000000\,\text{м/с}
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Given
n
1{,}5\,\text{безразм.}
c
300000000\,\text{м/с}
Find

v

  1. v=\frac{c}{n}v=\frac{300000000}{1{,}5}v=200000000\,\text{м/с}

    The refractive index shows the slowing down of light.

Answer200000000\,\text{м/с}

Position in this set: 14 from 24

Problem #123

Reflection and refraction of light

Refractive index of the medium 2. Speed of light in vacuum 300000000 m/s. Find the speed of light in the medium.

Answer options

  1. 300000000\,\text{м/с}
  2. 150000000\,\text{м/с}
  3. 75000000\,\text{м/с}
  4. 450000000\,\text{м/с}
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Given
n
2\,\text{безразм.}
c
300000000\,\text{м/с}
Find

v

  1. v=\frac{c}{n}v=\frac{300000000}{2}v=150000000\,\text{м/с}

    The refractive index shows the slowing down of light.

Answer150000000\,\text{м/с}

Position in this set: 15 from 24

Problem #124

Reflection and refraction of light

When passing from air to medium, the sine of the angle of incidence 0{,}6, sine of the angle of refraction 0{,}4. Find the speed of light in the medium; c = 300000000 m/s, air index is 1.

Answer options

  1. 400000000\,\text{м/с}
  2. 200000000\,\text{м/с}
  3. 100000000\,\text{м/с}
  4. 600000000\,\text{м/с}
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Given
a
0{,}6\,\text{безразм.}
b
0{,}4\,\text{безразм.}
c
300000000\,\text{м/с}
Find

v

  1. n=\frac{\sin\alpha}{\sin\beta}n=\frac{0{,}6}{0{,}4}n=1{,}5\,\text{безразм.}

    The law of refraction relates the sines of angles.

  2. v=\frac{c}{n}v=\frac{300000000}{1{,}5}v=200000000\,\text{м/с}

    We find the speed from the refractive index.

Answer200000000\,\text{м/с}

Position in this set: 16 from 24

Problem #125

Reflection and refraction of light

When passing from air to medium, the sine of the angle of incidence 0{,}8, sine of the angle of refraction 0{,}4. Find the speed of light in the medium; c = 300000000 m/s, air index is 1.

Answer options

  1. 300000000\,\text{м/с}
  2. 75000000\,\text{м/с}
  3. 150000000\,\text{м/с}
  4. 450000000\,\text{м/с}
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Given
a
0{,}8\,\text{безразм.}
b
0{,}4\,\text{безразм.}
c
300000000\,\text{м/с}
Find

v

  1. n=\frac{\sin\alpha}{\sin\beta}n=\frac{0{,}8}{0{,}4}n=2\,\text{безразм.}

    The law of refraction relates the sines of angles.

  2. v=\frac{c}{n}v=\frac{300000000}{2}v=150000000\,\text{м/с}

    We find the speed from the refractive index.

Answer150000000\,\text{м/с}

Position in this set: 17 from 24

Problem #126

Reflection and refraction of light

Light travels normally through layers of thickness 0{,}2 and 0{,}3 m with indicators 1{,}5 and 2. Find the total transit time in nanoseconds. c = 300000000 m/s.

Answer options

  1. 6\,\text{нс}
  2. 1{,}5\,\text{нс}
  3. 3\,\text{нс}
  4. 9\,\text{нс}
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Given
d_1
0{,}2\,\text{м}
d_2
0{,}3\,\text{м}
n_1
1{,}5\,\text{безразм.}
n_2
2\,\text{безразм.}
c
300000000\,\text{м/с}
Find

t_{\text{нс}}

  1. t_1=\frac{d_1n_1}{c}t_1=\frac{0{,}2 \cdot 1{,}5}{300000000}t_1=1\cdot 10^{-9}\,\text{с}

    The time of the first layer is equal to the thickness divided by c/n₁.

  2. t_2=\frac{d_2n_2}{c}t_2=\frac{0{,}3 \cdot 2}{300000000}t_2=2\cdot 10^{-9}\,\text{с}

    We calculate the second layer separately.

  3. t_{\text{нс}}=(t_1+t_2)10^9t_{\text{нс}}=(1\cdot 10^{-9} + 2\cdot 10^{-9}) \cdot 1000000000t_{\text{нс}}=3\,\text{нс}

    Add up the times and convert to nanoseconds.

Answer3\,\text{нс}

Position in this set: 18 from 24

Problem #127

Reflection and refraction of light

Light travels normally through layers of thickness 0{,}3 and 0{,}2 m with indicators 2 and 1{,}5. Find the total transit time in nanoseconds. c = 300000000 m/s.

Answer options

  1. 6\,\text{нс}
  2. 1{,}5\,\text{нс}
  3. 9\,\text{нс}
  4. 3\,\text{нс}
Show the walkthroughHide the walkthrough
Given
d_1
0{,}3\,\text{м}
d_2
0{,}2\,\text{м}
n_1
2\,\text{безразм.}
n_2
1{,}5\,\text{безразм.}
c
300000000\,\text{м/с}
Find

t_{\text{нс}}

  1. t_1=\frac{d_1n_1}{c}t_1=\frac{0{,}3 \cdot 2}{300000000}t_1=2\cdot 10^{-9}\,\text{с}

    The time of the first layer is equal to the thickness divided by c/n₁.

  2. t_2=\frac{d_2n_2}{c}t_2=\frac{0{,}2 \cdot 1{,}5}{300000000}t_2=1\cdot 10^{-9}\,\text{с}

    We calculate the second layer separately.

  3. t_{\text{нс}}=(t_1+t_2)10^9t_{\text{нс}}=(2\cdot 10^{-9} + 1\cdot 10^{-9}) \cdot 1000000000t_{\text{нс}}=3\,\text{нс}

    Add up the times and convert to nanoseconds.

Answer3\,\text{нс}

Position in this set: 19 from 24

Problem #128

Lenses and diffraction

Focal length of the collecting lens 0{,}2 m. Find the optical power.

Answer options

  1. 10\,\text{дптр}
  2. 5\,\text{дптр}
  3. 2{,}5\,\text{дптр}
  4. 15\,\text{дптр}
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Given
F
0{,}2\,\text{м}
Find

D

  1. D=\frac{1}{F}D=\frac{1}{0{,}2}D=5\,\text{дптр}

    For diopters, we take the focal length in meters.

Answer5\,\text{дптр}

Position in this set: 20 from 24

Problem #129

Lenses and diffraction

Focal length of the collecting lens 0{,}25 m. Find the optical power.

Answer options

  1. 8\,\text{дптр}
  2. 2\,\text{дптр}
  3. 4\,\text{дптр}
  4. 12\,\text{дптр}
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Given
F
0{,}25\,\text{м}
Find

D

  1. D=\frac{1}{F}D=\frac{1}{0{,}25}D=4\,\text{дптр}

    For diopters, we take the focal length in meters.

Answer4\,\text{дптр}

Position in this set: 21 from 24

Problem #130

Lenses and diffraction

Optical power of the collecting lens 4 diopter Find the focal length in centimeters.

Answer options

  1. 50\,\text{см}
  2. 12{,}5\,\text{см}
  3. 25\,\text{см}
  4. 75\,\text{см}
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Given
D
4\,\text{дптр}
Find

F_{\text{см}}

  1. F=\frac{1}{D}F=\frac{1}{4}F=0{,}25\,\text{м}

    The inverse optical power gives the focus in meters.

  2. F_{\text{см}}=100FF_{\text{см}}=0{,}25 \cdot 100F_{\text{см}}=25\,\text{см}

    Convert meters to centimeters.

Answer25\,\text{см}

Position in this set: 22 from 24

Problem #131

Lenses and diffraction

Optical power of the collecting lens 5 diopter Find the focal length in centimeters.

Answer options

  1. 40\,\text{см}
  2. 10\,\text{см}
  3. 60\,\text{см}
  4. 20\,\text{см}
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Given
D
5\,\text{дптр}
Find

F_{\text{см}}

  1. F=\frac{1}{D}F=\frac{1}{5}F=0{,}2\,\text{м}

    The inverse optical power gives the focus in meters.

  2. F_{\text{см}}=100FF_{\text{см}}=0{,}2 \cdot 100F_{\text{см}}=20\,\text{см}

    Convert meters to centimeters.

Answer20\,\text{см}

Position in this set: 23 from 24

Problem #132

Lenses and diffraction

Object height 3 cm located in 30 cm in front of a thin converging lens of focal length 20 cm. Find the magnitude of the real image height. Use 1/F = 1/d + 1/f and H/h = f/d, where d is the object distance and f the image distance.

Answer options

  1. 12\,\text{см}
  2. 3\,\text{см}
  3. 18\,\text{см}
  4. 6\,\text{см}
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Given
h
3\,\text{см}
d
30\,\text{см}
F
20\,\text{см}
Find

H

  1. f=\frac{dF}{d-F}f=\frac{30 \cdot 20}{(30 - 20)}f=60\,\text{см}

    The object is further than the focus, the image is real.

  2. |\Gamma|=\frac{f}{d}|\Gamma|=\frac{60}{30}|\Gamma|=2\,\text{раз}

    We calculate the linear increase modulus.

  3. H=h|\Gamma|H=3 \cdot 2H=6\,\text{см}

    The height module of the inverted image is required.

Answer6\,\text{см}

Position in this set: 24 from 24

Problem #133

Lenses and diffraction

Object height 4 cm located in 60 cm in front of a thin converging lens of focal length 20 cm. Find the magnitude of the real image height. Use 1/F = 1/d + 1/f and H/h = f/d, where d is the object distance and f the image distance.

Answer options

  1. 2\,\text{см}
  2. 4\,\text{см}
  3. 1\,\text{см}
  4. 6\,\text{см}
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Given
h
4\,\text{см}
d
60\,\text{см}
F
20\,\text{см}
Find

H

  1. f=\frac{dF}{d-F}f=\frac{60 \cdot 20}{(60 - 20)}f=30\,\text{см}

    The object is further than the focus, the image is real.

  2. |\Gamma|=\frac{f}{d}|\Gamma|=\frac{30}{60}|\Gamma|=0{,}5\,\text{раз}

    We calculate the linear increase modulus.

  3. H=h|\Gamma|H=4 \cdot 0{,}5H=2\,\text{см}

    The height module of the inverted image is required.

Answer2\,\text{см}

The theory behind these problems

Work through the topic if a problem will not give in.

Reflection and refraction of lightThe laws of reflection, refraction and total internal reflection.Lenses and diffractionThe lens equation, magnification, optical power and the diffraction grating.

Formulas by topic

Look up symbols and units of measurement in the reference.

Lens powerD · дптрPeriodT · сRefractive indexn · безразм.Distance and pathS, l · мLens magnificationΓ · безразм.Focal lengthF · м