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GRADE 8 PRACTICE

Physics Problems: Light phenomena, Grade 8

Light sources, propagation, reflection, refraction and lenses.

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 #217

Sources and properties of light

Light passes 600000 km at speed 300000 km/s. Find time distribution.

Answer options

  1. 2\,\text{с}
  2. 4\,\text{с}
  3. 1\,\text{с}
  4. 6\,\text{с}
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Given
S
600000\,\text{км}
c
300000\,\text{км/с}
Find

t

  1. t=\frac{S}{c}t=\frac{600000}{300000}t=2\,\text{с}

    We use the same units of travel distance and speed.

Answer2\,\text{с}

Position in this set: 2 from 24

Problem #218

Sources and properties of light

Light passes 900000 km at speed 300000 km/s. Find time distribution.

Answer options

  1. 6\,\text{с}
  2. 3\,\text{с}
  3. 1{,}5\,\text{с}
  4. 9\,\text{с}
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Given
S
900000\,\text{км}
c
300000\,\text{км/с}
Find

t

  1. t=\frac{S}{c}t=\frac{900000}{300000}t=3\,\text{с}

    We use the same units of travel distance and speed.

Answer3\,\text{с}

Position in this set: 3 from 24

Problem #219

Sources and properties of light

Spot source is located At a distance 20 cm from opaque disk diameter 4 cm. screen located At a distance 100 cm from source for disk. Find diameter shadows. Use resemblance: D/d = b/a.

Answer options

  1. 40\,\text{см}
  2. 20\,\text{см}
  3. 10\,\text{см}
  4. 60\,\text{см}
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Given
a
20\,\text{см}
d
4\,\text{см}
b
100\,\text{см}
Find

D

  1. k=\frac{b}{a}k=\frac{100}{20}k=5\,\text{раз}

    Comparing distances from one source.

  2. D=dkD=4 \cdot 5D=20\,\text{см}

    The straightness of the rays allows the use of similar triangles.

Answer20\,\text{см}

Position in this set: 4 from 24

Problem #220

Sources and properties of light

Spot source is located At a distance 30 cm from opaque disk diameter 6 cm. screen located At a distance 120 cm from source for disk. Find diameter shadows. Use resemblance: D/d = b/a.

Answer options

  1. 48\,\text{см}
  2. 12\,\text{см}
  3. 24\,\text{см}
  4. 72\,\text{см}
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Given
a
30\,\text{см}
d
6\,\text{см}
b
120\,\text{см}
Find

D

  1. k=\frac{b}{a}k=\frac{120}{30}k=4\,\text{раз}

    Comparing distances from one source.

  2. D=dkD=6 \cdot 4D=24\,\text{см}

    The straightness of the rays allows the use of similar triangles.

Answer24\,\text{см}

Position in this set: 5 from 24

Problem #221

Sources and properties of light

Spot source is located B 20 cm before subject height 5 cm. screen First located B 80 cm from source, then him move away more on 40 cm. On how much will increase height shadows? Use H/h = b/a.

Answer options

  1. 20\,\text{см}
  2. 5\,\text{см}
  3. 10\,\text{см}
  4. 30\,\text{см}
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Given
a
20\,\text{см}
h
5\,\text{см}
b
80\,\text{см}
x
40\,\text{см}
Find

\Delta H

  1. H_1=h\frac{b}{a}H_1=\frac{5 \cdot 80}{20}H_1=20\,\text{см}

    Calculate the initial height of the shadow.

  2. H_2=h\frac{b+x}{a}H_2=\frac{5 \cdot (80 + 40)}{20}H_2=30\,\text{см}

    We take into account the new distance to the screen.

  3. \Delta H=H_2-H_1\Delta H=30 - 20\Delta H=10\,\text{см}

    Finding the change in size.

Answer10\,\text{см}

Position in this set: 6 from 24

Problem #222

Sources and properties of light

Spot source is located B 25 cm before subject height 10 cm. screen First located B 100 cm from source, then him move away more on 50 cm. On how much will increase height shadows? Use H/h = b/a.

Answer options

  1. 40\,\text{см}
  2. 10\,\text{см}
  3. 60\,\text{см}
  4. 20\,\text{см}
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Given
a
25\,\text{см}
h
10\,\text{см}
b
100\,\text{см}
x
50\,\text{см}
Find

\Delta H

  1. H_1=h\frac{b}{a}H_1=\frac{10 \cdot 100}{25}H_1=40\,\text{см}

    Calculate the initial height of the shadow.

  2. H_2=h\frac{b+x}{a}H_2=\frac{10 \cdot (100 + 50)}{25}H_2=60\,\text{см}

    We take into account the new distance to the screen.

  3. \Delta H=H_2-H_1\Delta H=60 - 40\Delta H=20\,\text{см}

    Finding the change in size.

Answer20\,\text{см}

Position in this set: 7 from 24

Problem #223

Propagation of light

Light passes 600000 km at speed 300000 km/s. Find time distribution.

Answer options

  1. 4\,\text{с}
  2. 2\,\text{с}
  3. 1\,\text{с}
  4. 6\,\text{с}
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Given
S
600000\,\text{км}
c
300000\,\text{км/с}
Find

t

  1. t=\frac{S}{c}t=\frac{600000}{300000}t=2\,\text{с}

    We use the same units of travel distance and speed.

Answer2\,\text{с}

Position in this set: 8 from 24

Problem #224

Propagation of light

Light passes 900000 km at speed 300000 km/s. Find time distribution.

Answer options

  1. 6\,\text{с}
  2. 1{,}5\,\text{с}
  3. 3\,\text{с}
  4. 9\,\text{с}
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Given
S
900000\,\text{км}
c
300000\,\text{км/с}
Find

t

  1. t=\frac{S}{c}t=\frac{900000}{300000}t=3\,\text{с}

    We use the same units of travel distance and speed.

Answer3\,\text{с}

Position in this set: 9 from 24

Problem #225

Propagation of light

Spot source is located At a distance 20 cm from opaque disk diameter 4 cm. screen located At a distance 100 cm from source for disk. Find diameter shadows. Use resemblance: D/d = b/a.

Answer options

  1. 40\,\text{см}
  2. 10\,\text{см}
  3. 20\,\text{см}
  4. 60\,\text{см}
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Given
a
20\,\text{см}
d
4\,\text{см}
b
100\,\text{см}
Find

D

  1. k=\frac{b}{a}k=\frac{100}{20}k=5\,\text{раз}

    Comparing distances from one source.

  2. D=dkD=4 \cdot 5D=20\,\text{см}

    The straightness of the rays allows the use of similar triangles.

Answer20\,\text{см}

Position in this set: 10 from 24

Problem #226

Propagation of light

Spot source is located At a distance 30 cm from opaque disk diameter 6 cm. screen located At a distance 120 cm from source for disk. Find diameter shadows. Use resemblance: D/d = b/a.

Answer options

  1. 48\,\text{см}
  2. 12\,\text{см}
  3. 72\,\text{см}
  4. 24\,\text{см}
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Given
a
30\,\text{см}
d
6\,\text{см}
b
120\,\text{см}
Find

D

  1. k=\frac{b}{a}k=\frac{120}{30}k=4\,\text{раз}

    Comparing distances from one source.

  2. D=dkD=6 \cdot 4D=24\,\text{см}

    The straightness of the rays allows the use of similar triangles.

Answer24\,\text{см}

Position in this set: 11 from 24

Problem #227

Propagation of light

Spot source is located B 20 cm before subject height 5 cm. screen First located B 80 cm from source, then him move away more on 40 cm. On how much will increase height shadows? Use H/h = b/a.

Answer options

  1. 20\,\text{см}
  2. 5\,\text{см}
  3. 30\,\text{см}
  4. 10\,\text{см}
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Given
a
20\,\text{см}
h
5\,\text{см}
b
80\,\text{см}
x
40\,\text{см}
Find

\Delta H

  1. H_1=h\frac{b}{a}H_1=\frac{5 \cdot 80}{20}H_1=20\,\text{см}

    Calculate the initial height of the shadow.

  2. H_2=h\frac{b+x}{a}H_2=\frac{5 \cdot (80 + 40)}{20}H_2=30\,\text{см}

    We take into account the new distance to the screen.

  3. \Delta H=H_2-H_1\Delta H=30 - 20\Delta H=10\,\text{см}

    Finding the change in size.

Answer10\,\text{см}

Position in this set: 12 from 24

Problem #228

Propagation of light

Spot source is located B 25 cm before subject height 10 cm. screen First located B 100 cm from source, then him move away more on 50 cm. On how much will increase height shadows? Use H/h = b/a.

Answer options

  1. 20\,\text{см}
  2. 40\,\text{см}
  3. 10\,\text{см}
  4. 60\,\text{см}
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Given
a
25\,\text{см}
h
10\,\text{см}
b
100\,\text{см}
x
50\,\text{см}
Find

\Delta H

  1. H_1=h\frac{b}{a}H_1=\frac{10 \cdot 100}{25}H_1=40\,\text{см}

    Calculate the initial height of the shadow.

  2. H_2=h\frac{b+x}{a}H_2=\frac{10 \cdot (100 + 50)}{25}H_2=60\,\text{см}

    We take into account the new distance to the screen.

  3. \Delta H=H_2-H_1\Delta H=60 - 40\Delta H=20\,\text{см}

    Finding the change in size.

Answer20\,\text{см}

Position in this set: 13 from 24

Problem #229

Reflection of light and mirrors

The incident ray forms an angle with the surface of the mirror 60°. Find the angle of reflection relative to the normal.

Answer options

  1. 60\,\text{°}
  2. 15\,\text{°}
  3. 30\,\text{°}
  4. 90\,\text{°}
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Given
a
60\,\text{°}
Find

\beta

  1. \beta=90^\circ-a\beta=90 - 60\beta=30\,\text{°}

    The angles of incidence and reflection are measured from the normal.

Answer30\,\text{°}

Position in this set: 14 from 24

Problem #230

Reflection of light and mirrors

The incident ray forms an angle with the surface of the mirror 45°. Find the angle of reflection relative to the normal.

Answer options

  1. 90\,\text{°}
  2. 22{,}5\,\text{°}
  3. 135\,\text{°}
  4. 45\,\text{°}
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Given
a
45\,\text{°}
Find

\beta

  1. \beta=90^\circ-a\beta=90 - 45\beta=45\,\text{°}

    The angles of incidence and reflection are measured from the normal.

Answer45\,\text{°}

Position in this set: 15 from 24

Problem #231

Reflection of light and mirrors

The beam forms an angle with the surface of a flat mirror 60°. Find the angle between the incident and reflected rays.

Answer options

  1. 120\,\text{°}
  2. 30\,\text{°}
  3. 180\,\text{°}
  4. 60\,\text{°}
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Given
a
60\,\text{°}
Find

\gamma

  1. \alpha=90^\circ-a\alpha=90 - 60\alpha=30\,\text{°}

    We move from the angle with the surface to the angle with the normal.

  2. \gamma=2\alpha\gamma=2 \cdot 30\gamma=60\,\text{°}

    The angles on both sides of the normal are the same.

Answer60\,\text{°}

Position in this set: 16 from 24

Problem #232

Reflection of light and mirrors

The beam forms an angle with the surface of a flat mirror 55°. Find the angle between the incident and reflected rays.

Answer options

  1. 70\,\text{°}
  2. 140\,\text{°}
  3. 35\,\text{°}
  4. 210\,\text{°}
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Given
a
55\,\text{°}
Find

\gamma

  1. \alpha=90^\circ-a\alpha=90 - 55\alpha=35\,\text{°}

    We move from the angle with the surface to the angle with the normal.

  2. \gamma=2\alpha\gamma=2 \cdot 35\gamma=70\,\text{°}

    The angles on both sides of the normal are the same.

Answer70\,\text{°}

Position in this set: 17 from 24

Problem #233

Reflection of light and mirrors

The 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?

Answer options

  1. 30\,\text{см}
  2. 60\,\text{см}
  3. 15\,\text{см}
  4. 90\,\text{см}
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Given
d
80\,\text{см}
x
15\,\text{см}
Find

\Delta D

  1. D_1=2dD_1=2 \cdot 80D_1=160\,\text{см}

    The image is symmetrical to the object relative to the mirror.

  2. D_2=2(d-x)D_2=2 \cdot (80 - 15)D_2=130\,\text{см}

    Both symmetrical parts of the distance decrease.

  3. \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}

Position in this set: 18 from 24

Problem #234

Reflection of light and mirrors

The 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?

Answer options

  1. 80\,\text{см}
  2. 40\,\text{см}
  3. 20\,\text{см}
  4. 120\,\text{см}
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Given
d
100\,\text{см}
x
20\,\text{см}
Find

\Delta D

  1. D_1=2dD_1=2 \cdot 100D_1=200\,\text{см}

    The image is symmetrical to the object relative to the mirror.

  2. D_2=2(d-x)D_2=2 \cdot (100 - 20)D_2=160\,\text{см}

    Both symmetrical parts of the distance decrease.

  3. \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}

Position in this set: 19 from 24

Problem #235

Refraction and lenses

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

Answer options

  1. 10\,\text{дптр}
  2. 2{,}5\,\text{дптр}
  3. 15\,\text{дптр}
  4. 5\,\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 #236

Refraction and lenses

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

Answer options

  1. 4\,\text{дптр}
  2. 8\,\text{дптр}
  3. 2\,\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 #237

Refraction and lenses

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

Answer options

  1. 25\,\text{см}
  2. 50\,\text{см}
  3. 12{,}5\,\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 #238

Refraction and lenses

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

Answer options

  1. 40\,\text{см}
  2. 20\,\text{см}
  3. 10\,\text{см}
  4. 60\,\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 #239

Refraction and lenses

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. 6\,\text{см}
  3. 3\,\text{см}
  4. 18\,\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 #240

Refraction and lenses

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. 4\,\text{см}
  2. 1\,\text{см}
  3. 2\,\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.

Sources and properties of lightNatural and artificial sources, effects of light and its energy.Propagation of lightLight rays, straight-line propagation, umbra and penumbra.Reflection of light and mirrorsThe law of reflection and the construction of an image in a plane mirror.Refraction and lensesChanges in light direction, lens focus and image construction.