GRADE 9 PRACTICE
Physics Problems: Light phenomena, Grade 9
Geometric and wave optics, lenses, color and spectra.
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.
Problem #40
Light propagation and reflectionWhat concept does the description correspond to: “The incident ray, the reflected ray and the normal lie in the same plane, and the angles of incidence and reflection are equal.”?
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- \text{Description}
- \text{The incident ray, reflected ray and normal lie in one plane, and the angles of incidence and reflection are equal.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Angles are measured from the normal to the surface, not from the surface itself.
- \text{Let's formulate conclusion}\text{Concept} = \text{Law of reflection}
Answer\text{Concept} = \text{Law of reflection}
Problem #41
Light propagation and reflectionWhich concept corresponds to the description: “A plane mirror gives a virtual direct image of equal size at the same distance behind the mirror.”?
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- \text{Description}
- \text{A plane mirror forms a virtual, upright image of equal size at the same distance behind the mirror.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Construct the image by extending reflected rays backward until their extensions meet behind the mirror.
- \text{Let's formulate conclusion}\text{Concept} = \text{Image in a plane mirror}
Answer\text{Concept} = \text{Image in a plane mirror}
Problem #42
Reflection of lightThe beam hits the mirror at an angle of 35° to the normal. What is the angle of reflection?
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- \alpha
- 35^{\circ}
\beta - ?
- \alpha=\beta\beta=35^{\circ}\alpha = 35^{\circ}
Answer35^{\circ}
Problem #43
Refraction and total internal reflectionWhat concept does the description correspond to: “The absolute indicator is equal to the ratio of the speed of light in a vacuum to the speed of light in a medium.”?
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- \text{Description}
- \text{Absolute refractive index is the ratio of vacuum light speed to light speed in the medium.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
A higher refractive index means lower light speed in the medium and usually stronger bending towards the normal.
- \text{Let's formulate conclusion}\text{Concept} = \text{Refractive index}
Answer\text{Concept} = \text{Refractive index}
Problem #44
Refraction and total internal reflectionWhat concept does the description correspond to: “When moving between media with different refractive indices, the speed of light changes. The direction of the beam changes with oblique incidence; when falling normally, it is preserved.”?
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- \text{Description}
- \text{Light speed changes between media of different refractive indices. A ray changes direction at oblique incidence; at normal incidence it keeps its direction.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
On entering an optically denser medium, a ray usually bends towards the normal.
- \text{Let's formulate conclusion}\text{Concept} = \text{Refraction of light}
Answer\text{Concept} = \text{Refraction of light}
Problem #45
Refraction of lightLight propagates in a transparent medium at a speed of 2·10⁸ m/s. The speed of light in vacuum is 3·10⁸ m/s. Find the indicator.
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- c
- 3\cdot10^8\,\text{м/с}
- v
- 2\cdot10^8\,\text{м/с}
n - ?
- n=\frac{c}{v}n=\frac{3\cdot10^8}{2\cdot10^8}n = 1{,}5
Answer1{,}5\,\text{безразм.}
Problem #46
Lenses, the eye and optical instrumentsWhat concept does the description correspond to: “A converging lens turns a parallel beam into a converging one, and a diverging lens makes it divergent.”?
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- \text{Description}
- \text{A converging lens makes a parallel beam converge; a diverging lens makes it spread out.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
The type of lens in the medium is determined by the shape of the surfaces and the relative refractive index.
- \text{Let's formulate conclusion}\text{Concept} = \text{Converging and diverging lenses}
Answer\text{Concept} = \text{Converging and diverging lenses}
Problem #47
Lenses, the eye and optical instrumentsWhat concept does the description correspond to: “The image is found by the intersection of rays or their continuations after passing through the lens.”?
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- \text{Description}
- \text{Locate an image where rays, or their extensions, intersect after passing through the lens.}
\text{name concepts} - ?
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A converging lens can form a real or virtual image; a diverging lens usually forms a smaller virtual image.
- \text{Let's formulate conclusion}\text{Concept} = \text{Constructing an image with a lens}
Answer\text{Concept} = \text{Constructing an image with a lens}
Problem #48
LensesThe focal length of the collecting lens is 0.25 m. Find its optical power.
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- F
- 0{,}25\,\text{м}
D - ?
- D=\frac{1}{F}D=\frac{1}{0{,}25}\,\text{дптр}D = 4\,\text{дптр}
Answer4\,\text{дптр}
Problem #49
Interference, diffraction and spectraWhat concept does the description correspond to: “Stable redistribution of intensity upon superposition of matched light waves.”?
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- \text{Description}
- \text{A stable redistribution of intensity when coherent light waves overlap.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
At some points the waves strengthen each other, at others they weaken each other, forming an alternation of maxima and minima.
- \text{Let's formulate conclusion}\text{Concept} = \text{Interference}
Answer\text{Concept} = \text{Interference}
Problem #50
Interference, diffraction and spectraWhat concept does the description correspond to: “The deviation of waves from straight-line propagation near obstacles and holes, noticeable at comparable sizes.”?
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- \text{Description}
- \text{Departure from straight-line wave propagation near obstacles and openings, noticeable when their sizes are comparable to wavelength.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
A diffraction grating separates light by wavelength and is used to observe spectra.
- \text{Let's formulate conclusion}\text{Concept} = \text{Diffraction}
Answer\text{Concept} = \text{Diffraction}
Problem #51
SpectrumThe wavelength of light in a vacuum is 600 nm. Find the frequency at c = 3·10⁸ m/s.
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- \lambda
- 600\,\text{нм}=6\cdot10^{-7}\,\text{м}
- c
- 3\cdot10^8\,\text{м/с}
\nu - ?
- c=\lambda\nu
- \nu=\frac{c}{\lambda}\nu=\frac{3\cdot10^8}{6\cdot10^{-7}}\,\text{Гц}\nu = 5\cdot10^{14}\,\text{Гц}
Answer5\cdot10^{14}\,\text{Гц}
Problem #142
Light propagation and reflectionThe incident ray forms an angle with the surface of the mirror 60°. Find the angle of reflection relative to the normal.
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- a
- 60\,\text{°}
\beta - ?
- \beta=90^\circ-a\beta=90 - 60\beta=30\,\text{°}
The angles of incidence and reflection are measured from the normal.
Answer30\,\text{°}
Problem #143
Light propagation and reflectionThe incident ray forms an angle with the surface of the mirror 45°. Find the angle of reflection relative to the normal.
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- a
- 45\,\text{°}
\beta - ?
- \beta=90^\circ-a\beta=90 - 45\beta=45\,\text{°}
The angles of incidence and reflection are measured from the normal.
Answer45\,\text{°}
Problem #144
Light propagation and reflectionThe beam forms an angle with the surface of a flat mirror 60°. Find the angle between the incident and reflected rays.
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- a
- 60\,\text{°}
\gamma - ?
- \alpha=90^\circ-a\alpha=90 - 60\alpha=30\,\text{°}
We move from the angle with the surface to the angle with the normal.
- \gamma=2\alpha\gamma=2 \cdot 30\gamma=60\,\text{°}
The angles on both sides of the normal are the same.
Answer60\,\text{°}
Problem #145
Light propagation and reflectionThe beam forms an angle with the surface of a flat mirror 55°. Find the angle between the incident and reflected rays.
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- a
- 55\,\text{°}
\gamma - ?
- \alpha=90^\circ-a\alpha=90 - 55\alpha=35\,\text{°}
We move from the angle with the surface to the angle with the normal.
- \gamma=2\alpha\gamma=2 \cdot 35\gamma=70\,\text{°}
The angles on both sides of the normal are the same.
Answer70\,\text{°}
Problem #146
Light propagation and reflectionThe item is in 80 cm in front of a flat mirror. He approaches the stationary mirror on 15 cm. By how many centimeters will the distance between the object and the image decrease?
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- d
- 80\,\text{см}
- x
- 15\,\text{см}
\Delta D - ?
- D_1=2dD_1=2 \cdot 80D_1=160\,\text{см}
The image is symmetrical to the object relative to the mirror.
- D_2=2(d-x)D_2=2 \cdot (80 - 15)D_2=130\,\text{см}
Both symmetrical parts of the distance decrease.
- \Delta D=D_1-D_2\Delta D=160 - 130\Delta D=30\,\text{см}
A decrease is equal to twice the movement of an object.
Answer30\,\text{см}
Problem #147
Light propagation and reflectionThe item is in 100 cm in front of a flat mirror. He approaches the stationary mirror on 20 cm. By how many centimeters will the distance between the object and the image decrease?
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- d
- 100\,\text{см}
- x
- 20\,\text{см}
\Delta D - ?
- D_1=2dD_1=2 \cdot 100D_1=200\,\text{см}
The image is symmetrical to the object relative to the mirror.
- D_2=2(d-x)D_2=2 \cdot (100 - 20)D_2=160\,\text{см}
Both symmetrical parts of the distance decrease.
- \Delta D=D_1-D_2\Delta D=200 - 160\Delta D=40\,\text{см}
A decrease is equal to twice the movement of an object.
Answer40\,\text{см}
Problem #148
Refraction and total internal reflectionRefractive index of the medium 1{,}5. Speed of light in vacuum 300000000 m/s. Find the speed of light in the medium.
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- n
- 1{,}5\,\text{безразм.}
- c
- 300000000\,\text{м/с}
v - ?
- 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}
Problem #149
Refraction and total internal reflectionRefractive index of the medium 2. Speed of light in vacuum 300000000 m/s. Find the speed of light in the medium.
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- n
- 2\,\text{безразм.}
- c
- 300000000\,\text{м/с}
v - ?
- 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}
Problem #150
Refraction and total internal reflectionWhen 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.
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- a
- 0{,}6\,\text{безразм.}
- b
- 0{,}4\,\text{безразм.}
- c
- 300000000\,\text{м/с}
v - ?
- 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.
- 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}
Problem #151
Refraction and total internal reflectionWhen 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.
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- a
- 0{,}8\,\text{безразм.}
- b
- 0{,}4\,\text{безразм.}
- c
- 300000000\,\text{м/с}
v - ?
- 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.
- 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}
Problem #152
Refraction and total internal reflectionLight 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.
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- 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}
t_{\text{нс}} - ?
- 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.
- 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.
- 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}
Problem #153
Refraction and total internal reflectionLight 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.
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- 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}
t_{\text{нс}} - ?
- 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.
- 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.
- 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}
Problem #154
Lenses, the eye and optical instrumentsFocal length of the collecting lens 0{,}2 m. Find the optical power.
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- F
- 0{,}2\,\text{м}
D - ?
- 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}
Problem #155
Lenses, the eye and optical instrumentsFocal length of the collecting lens 0{,}25 m. Find the optical power.
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- F
- 0{,}25\,\text{м}
D - ?
- 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}
Problem #156
Lenses, the eye and optical instrumentsOptical power of the collecting lens 4 diopter Find the focal length in centimeters.
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- D
- 4\,\text{дптр}
F_{\text{см}} - ?
- F=\frac{1}{D}F=\frac{1}{4}F=0{,}25\,\text{м}
The inverse optical power gives the focus in meters.
- F_{\text{см}}=100FF_{\text{см}}=0{,}25 \cdot 100F_{\text{см}}=25\,\text{см}
Convert meters to centimeters.
Answer25\,\text{см}
Problem #157
Lenses, the eye and optical instrumentsOptical power of the collecting lens 5 diopter Find the focal length in centimeters.
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- D
- 5\,\text{дптр}
F_{\text{см}} - ?
- F=\frac{1}{D}F=\frac{1}{5}F=0{,}2\,\text{м}
The inverse optical power gives the focus in meters.
- F_{\text{см}}=100FF_{\text{см}}=0{,}2 \cdot 100F_{\text{см}}=20\,\text{см}
Convert meters to centimeters.
Answer20\,\text{см}
Problem #158
Lenses, the eye and optical instrumentsObject 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.
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- h
- 3\,\text{см}
- d
- 30\,\text{см}
- F
- 20\,\text{см}
H - ?
- 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.
- |\Gamma|=\frac{f}{d}|\Gamma|=\frac{60}{30}|\Gamma|=2\,\text{раз}
We calculate the linear increase modulus.
- H=h|\Gamma|H=3 \cdot 2H=6\,\text{см}
The height module of the inverted image is required.
Answer6\,\text{см}
Problem #159
Lenses, the eye and optical instrumentsObject 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.
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- h
- 4\,\text{см}
- d
- 60\,\text{см}
- F
- 20\,\text{см}
H - ?
- 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.
- |\Gamma|=\frac{f}{d}|\Gamma|=\frac{30}{60}|\Gamma|=0{,}5\,\text{раз}
We calculate the linear increase modulus.
- H=h|\Gamma|H=4 \cdot 0{,}5H=2\,\text{см}
The height module of the inverted image is required.
Answer2\,\text{см}
Problem #160
Interference, diffraction and spectraWave path difference 1200 nm, wavelength 600 nm. How many wavelengths is the path difference equal to?
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- d
- 1200\,\text{нм}
- l
- 600\,\text{нм}
N - ?
- N=\frac{\Delta}{\lambda}N=\frac{1200}{600}N=2\,\text{длин волн}
An integer number of wavelengths corresponds to the maximum for common-mode sources.
Answer2\,\text{длин волн}
Problem #161
Interference, diffraction and spectraWave path difference 1800 nm, wavelength 450 nm. How many wavelengths is the path difference equal to?
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- d
- 1800\,\text{нм}
- l
- 450\,\text{нм}
N - ?
- N=\frac{\Delta}{\lambda}N=\frac{1800}{450}N=4\,\text{длин волн}
An integer number of wavelengths corresponds to the maximum for common-mode sources.
Answer4\,\text{длин волн}
Problem #162
Interference, diffraction and spectraApplied to 1 mm of grating 500 strokes. Find its period in micrometers.
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- N
- 500\,\text{штрихов/мм}
D - ?
- d=\frac{1}{N}d=\frac{1}{500}d=0{,}002\,\text{мм}
The period is equal to the distance between adjacent strokes.
- D=1000dD=0{,}002 \cdot 1000D=2\,\text{мкм}
Convert millimeters to micrometers.
Answer2\,\text{мкм}
Problem #163
Interference, diffraction and spectraApplied to 1 mm of grating 250 strokes. Find its period in micrometers.
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- N
- 250\,\text{штрихов/мм}
D - ?
- d=\frac{1}{N}d=\frac{1}{250}d=0{,}004\,\text{мм}
The period is equal to the distance between adjacent strokes.
- D=1000dD=0{,}004 \cdot 1000D=4\,\text{мкм}
Convert millimeters to micrometers.
Answer4\,\text{мкм}
Problem #164
Interference, diffraction and spectraThe grille has 500 strokes per 1 mm. Maximum order 2 observed at the sine of the angle 0{,}6. Find the wavelength in nanometers. Use d sin θ = kλ.
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- N
- 500\,\text{штрихов/мм}
- k
- 2\,\text{порядок}
- a
- 0{,}6\,\text{безразм.}
R - ?
- d=\frac{10^{-3}}{N}d=\frac{0{,}001}{500}d=0{,}000002\,\text{м}
Convert the lattice period to SI.
- \lambda=\frac{d\sin\theta}{k}\lambda=\frac{0{,}000002 \cdot 0{,}6}{2}\lambda=6\cdot 10^{-7}\,\text{м}
We express the wavelength from the maximum condition.
- R=\lambda\cdot10^9R=6\cdot 10^{-7} \cdot 1000000000R=600\,\text{нм}
Let's return to optical units.
Answer600\,\text{нм}
Problem #165
Interference, diffraction and spectraThe grille has 400 strokes per 1 mm. Maximum order 2 observed at the sine of the angle 0{,}4. Find the wavelength in nanometers. Use d sin θ = kλ.
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- N
- 400\,\text{штрихов/мм}
- k
- 2\,\text{порядок}
- a
- 0{,}4\,\text{безразм.}
R - ?
- d=\frac{10^{-3}}{N}d=\frac{0{,}001}{400}d=0{,}0000025\,\text{м}
Convert the lattice period to SI.
- \lambda=\frac{d\sin\theta}{k}\lambda=\frac{0{,}0000025 \cdot 0{,}4}{2}\lambda=5\cdot 10^{-7}\,\text{м}
We express the wavelength from the maximum condition.
- R=\lambda\cdot10^9R=5\cdot 10^{-7} \cdot 1000000000R=500\,\text{нм}
Let's return to optical units.
Answer500\,\text{нм}
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