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Spectrum : Selina ICSE Concise Physics for Class 10 Physics Chapter 6

Selina ICSE Concise Physics  for Class 10 Physics
 Chapter 6 Spectrum
Exercise 6(A)

1.Name three factors on which the deviation produced by a prism depends and state how does it depend on the factors stated by you.


Solution 1.

The deviation produced by the prism depends on the following four factors:

  1. The angle of incidence – As the angle of incidence increases, first the angle of deviation decreases and reaches to a minimum value for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
  2. The material of prism (i.e., on refractive index) – For a given angle of incidence, the prism with a higher refractive index produces a greater deviation than the prism which has a lower refractive index.
  3. Angle of prism- Angle of deviation increases with the increase in the angle of prism.
  4. The colour or wavelength of light used- Angle of deviation increases with the decrease in wavelength of light.
2.How does the deviation produced by a triangular prism depend on the colour (or wavelength) of light incident on it? 

Solution 2.

The deviation caused by a prism increases with the decrease in the wavelength of light incident on it.

3.

Solution 3.

Speed of light increases with increase in the wavelength.

4.Which colour of white light travels (a) fastest  (b) slowest, in glass?

Solution 4.

Red colour travels fastest and Blue colour travels slowest in glass.

5.Name the subjective property of light related to its wavelength.

Solution 5.

Colour of light is related to its wavelength.

6.What is the range of wavelength of the spectrum of white light in (i) Selina Solutions Concise Physics Class 10 Chapter 6 - 1  and (ii) nm?

Solution 6.

(i) 4000 Γ… to 8000 Γ…
(ii) 400 nm to 800 nm

7.(a) Write the approximate wavelengths for (i) blue and (ii) red light.


Solution 7.

(i) For blue light, approximate wavelength = 4800 Γ…
(ii) For red light, approximate wavelength = 8000 Γ…
8.Write the seven prominent colours present in white light in the order of increasing wavelength

Solution 8.

Seven prominent colours of the white light spectrum in order of their increasing frequencies:
Red, Orange, Yellow, Green, Blue, Indigo, Violet

9.Name four colours of the spectrum of white light which have wavelength longer than blue light.

Solution 9.

Green, Yellow orange and red have wavelength longer than blue light.

10.Which colour of the white light is deviated by a glass prism (i) the most and, (ii) the least?

Solution 10.

A glass prism deviates the violet light most and the red light least.

11.The wavelengths for the light of red and blue colours are nearly 7.8 x 10-7m and 4.8 x 10-7 m respectively.

(a)Which colour has the greater speed in vacuum?

(b)Which colour has the greater speed in glass?

Solution 11.

(a) In vacuum, both have the same speeds.
(b) In glass, red light has a greater speed.

12.Define the term dispersion of light.

Solution 12.

The phenomenon of splitting of white light by a prism into its constituent colours is known as dispersion of light.

13.Explain the cause of dispersion of white light through a prism.

Solution 13.

When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism. Thus the cause of dispersion is the change in speed of light with wavelength or frequency.


14.Explain briefly, with the help of a neat labelled diagram, how does white light gets dispersed by a prism.
On which surface of prism, there is both the dispersion and deviation of light, and on which surface of prism, there is only the deviation of light?

Solution 14.


When white light is incident on the first surface of a prism and enters in glass, light of different colours due to different speeds in glass, is refracted or deviated through different angles. Thus the dispersion of white light into its constituent colours takes place at the first surface of prism.

Selina Concise Physics Class 10 ICSE Solutions Chapter 6 - Spectrum
On the second surface, only refraction takes place and different colours are deviated through different angles. As a result, the colours get further separated on refraction at the second surface (violet being deviated the most and red the least).

15.What do you understand by the term spectrum?

Solution 15.

The colour band obtained on a screen on passing white light through a prism is called the spectrum.

16.A ray of white light is passed through a glass prism and spectrum is obtained on a screen.

(a)Name the seven colours of the spectrum in order.

(b)Do the colours have the same width in the spectrum?

(c)Which of the colour of the spectrum of white light deviates (i) the most? (ii) the least?

Solution 16.

(a) Violet, Indigo, Blue, Green, Yellow, Orange, Red.
(b) No, different colours have different widths in the spectrum.
(c) (i) Violet colour is deviated the most. (ii) Red colour is deviated the least.

17.The diagram shown below shows the path taken by a narrow beam of yellow monochromatic light passing through an equiangular glass prism. If the yellow light is replaced by a narrow beam of white light incident at the same angle, draw another diagram to show the passage of the beam through the prism and label it to show the effect of prism on the white light.

Selina Solutions Concise Physics Class 10 Chapter 6 - 9

Solution 17.
  

Selina Solutions Concise Physics Class 10 Chapter 6 - 10

18.Figure shows a thin beam of white light from a source S striking on one face of a prism.

Selina Solutions Concise Physics Class 10 Chapter 6 - 11


Solution 18.


  • Constituent colours of white light are seen on the screen after dispersion through the prism.
Selina Solutions Concise Physics Class 10 Chapter 6 - 12
  • When a slit is introduced in between the prism and screen to pass only the light of green colour, only green light is observed on the screen.
  • From the observation, we conclude that prism itself produces no colour.
19.(a) A beam of monochromatic light undergoes minimum deviation through an equiangular prism, how does the beam pass through the prism, with respect to its base?

(b) If white light is used in same way as in part (a) above, what change is expected in the emergent beam?

(c) What conclusion do you draw about the nature of white light in part (b)?

Solution 19.

  • If a monochromatic beam of light undergoes minimum deviation through an equi-angular prism, then the beam passes parallel to the base of prism.
  • White light splits into its constituent colours i.e., spectrum is formed.
  • We conclude that white light is polychromatic.

1(MCQ).When a white light ray falls on a prism, the ray at its first surface suffers:

(a) no refraction

(b) only dispersion

(c) only deviation

(d) both deviation and dispersion

Solution 1 (MCQ).

Both deviation and dispersion.
Hint: When a white light ray falls on the first surface of a prism, light rays of different colours due to their different speeds in glass get refracted (or deviated) through different angles. Thus, the dispersion of white light into its constituent colours takes place at the first surface of prism.

2(MCQ).In the spectrum of white light by a prism, the colour at the extreme end opposite to the base of prism is:

(a) violet

(b) yellow

(c) red

(d) blue

Solution 2 (MCQ).

The colour of the extreme end opposite to the base of the prism is red.
Hint: The angle of deviation decreases with the increase in wavelength of light for a given angle of incidence. Since the red light has greatest wavelength, it gets deviated the least and is seen on the extreme end opposite to the base of prism.

Numericals

1.Calculate the frequency of yellow light of wavelength 550 nm. The speed of light is 3 × 108 m s-1.

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Chapter 6 - Spectrum

2.The frequency range of visible light is from 3.75 × 1014 Hz to 7.5 × 1014 Hz. Calculate its wavelength range. Take speed of light = 3 × 108 m s-1
  
Solution 2.
  
Selina Concise Physics Class 10 ICSE Solutions Chapter 6 - Spectrum

Exercise 6(B)

1.(a) Give a list of at least five radiations, in the order of their increasing wavelength, which make up the complete electromagnetic spectrum.

(b) Name the radiation mentioned by you in part (a) which has the highest penetrating power.

Solution 1.

(a) Five radiations in the order of their increasing frequencies are:
Infrared waves, Visible light, Ultraviolet, X-rays and Gamma rays.
(b) Gamma rays have the highest penetrating power.

2.(a) Arrange the following radiations in the order of their increasing wavelength:

     X-rays, infrared rays, radio waves, gamma rays and micro waves.


(b) Name the radiation which is used for satellite communication?

Solution 2.

(a) Gamma rays, X-rays, infrared rays, micro waves, radio waves.
(b) Microwave is used for satellite communication.

3.A wave has a wavelength of 10-3 nm (a) Name the wave (b) State its one property different from light.

Solution 3.

(a) Gamma ray.
(b) Gamma rays have strong penetrating power.

4.(a) Name the high energetic invisible electromagnetic wave which helps in the study of structure of crystals.

(b) State one more use of the wave named in part (a).

Solution 4.

(a) X-rays are used in the study of crystals.
(b) It is also used to detect fracture in bones.

5.Give the range of wavelength of the electromagnetic waves visible to us.

Solution 5.

4000 Γ… to 8000 Γ…

6.

Solution 6.

(i) Infrared
(ii) Ultraviolet

7.

Solution 7.

The part of spectrum beyond the red and the violet ends is called the invisible spectrum as our eyes do not respond to the spectrum beyond the red and the violet extremes.

8.Name the radiation which can be detected by (a) a thermopile (b) a solution of silver chloride.

Solution 8.

(a) infrared radiation
(b) ultra violet radiation

9.State the approximate range of wavelength associated with (a) the ultraviolet rays, (b) the visible light, and (c) infrared rays.

Solution 9.

(i) Ultraviolet rays-wavelength range 100Γ… to 4000Γ…

(ii) Visible light-wavelength range 4000Γ… to 8000Γ…

(iii) Infrared radiations-wavelength range 8000Γ… to 107Γ…

10.Name the radiations of wavelength just (a) longer than 8 × 10-7 m, (b) shorter than 4 × 10-7 m

Solution 10.

(i) Infrared radiations are longer than 8 x 10-7m.
(ii) ultraviolet radiations are shorter than 4 x 10-7 m.

11.Give one use each of (a) microwaves, (b) ultraviolet radiations, (c) infrared radiations, and (d) gamma rays.

Solution 11.

(i) Microwaves are used for satellite communication.
(ii) Ultraviolet radiations are used for detecting the purity of gems, eggs, ghee etc.

(iii) Infrared radiations are used in remote control of television and other gadgets.
(iv) Gamma rays are used in medical science to kill cancer cells.

12.Two waves A and B have wavelength 0.01 Selina Solutions Concise Physics Class 10 Chapter 6 - 33and 9000 Selina Solutions Concise Physics Class 10 Chapter 6 - 34 respectively.
(a) Name the two waves.
(b) Compare the speeds of these waves when they travel in vacuum.

Solution 12.

(a) A- Gamma rays, B-infrared radiations
(b) Ratio of speeds of these waves in vacuum is 1:1 as all electromagnetic waves travel with the speed of light in vacuum.

13.Name two sources, each of infrared radiations and ultraviolet radiations.

Solution 13.

All heated bodies such as a heated iron ball, flame, fire etc., are the sources of infrared radiations.
The electric arc and sparks give ultraviolet radiations.

14.What are infrared radiations? How are they detected? State one use of these radiations.

Solution 14.

Infrared radiations are the electromagnetic waves of wavelength in the range of 8000Γ… to 107Γ….

Detection: If a thermometer with a blackened bulb is moved from the violet end towards the red end, it is observed that there is a slow rise in temperature, but when it is moved beyond the red region, a rapid rise in temperature is noticed. It means that the portion of spectrum beyond the red end has certain radiations which produce a strong heating effect, but they are not visible. These radiations are called the infrared radiations.
Use: The infrared radiations are used for therapeutic purposes by doctors.

15.What are ultraviolet radiations? How are they detected? State one use of these radiations.

Solution 15..

The electromagnetic radiations of wavelength from 100Γ… to 4000Γ… are called the ultraviolet radiations.

Detection: If the different radiations from the red part of the spectrum to the violet end and beyond it, are made incident on the silver-chloride solution, it is observed that from the red to the violet end, the solution remains unaffected. However just beyond the violet end, it first turns violet and finally it becomes dark brown. Thus there exist certain radiations beyond the violet end of the spectrum, which are chemically more active than visible light, called ultraviolet radiations.
Use: Ultraviolet radiations are used for sterilizing purposes.

16.Name three properties of ultraviolet radiations which are similar to the visible light.

Solution 16.

  • Ultraviolet radiations travel in a straight line with a speed of 3 x 108 m in air (or vacuum).
  • They obey the laws of reflection and refraction.
  • They affect the photographic plate.
17.

Solution 17.

  • Ultraviolet radiations produce fluorescence on striking a zinc sulphide screen.
  • They cause health hazards like cancer on the body.
18.Mention three properties of infrared radiations similar to the visible light.

Solution 18.

  • Infrared radiations travel in straight line as light does, with a speed equal to 3 x 108m/s in vacuum.
  • They obey the laws of reflection and refraction.
  • They do not cause fluorescence on zinc sulphide screen.

19.Give reason for the following:

(i) Infrared radiations are used for photography in fog

(ii) Infrared radiations are used for signals during the war.

(iii)The photographic darkrooms are provided with infrared lamps.

(iv)A rock salt prism is used instead of a glass prism to obtain the infrared spectrum.

(v)A quartz prism is required for obtaining the spectrum of the ultraviolet light.

(vi)Ultraviolet bulbs have a quartz envelope instead of glass.

Solution 19.


  1. Infrared radiations are used in photography in fog because they are not much scattered by the atmosphere, so they can penetrate appreciably through it.
  2. Infrared radiations are used as signals during the war as they are not visible and they are not absorbed much in the medium.
  3. Infrared lamps are used in dark rooms for developing photographs since they do not affect the photographic film chemically, but they provide some visibility.
  4. Infrared spectrum can be obtained only with the help of a rock-salt prism since the rock-salt prism does not absorb infrared radiations whereas a glass prism absorbs them.
  5. A quartz prism is used to obtain the spectrum of the ultraviolet radiations as they are not absorbed by quartz, whereas ordinary glass absorbs the ultraviolet light.
  6. Ultraviolet bulbs have a quartz envelope instead of glass as they are not absorbed by quartz, whereas ordinary glad absorbs the ultraviolet light.

1(MCQ).The most energetic electromagnetic radiations are:

(a) microwaves

(b) ultraviolet waves

(c) X- rays

(d) gamma rays

Solution 1 (MCQ).

Gamma rays

2(MCQ).The source of ultraviolet light is:

(a) electric bulb

(b) red hot iron ball

(c) sodium vapour lamp

(d) carbon arc-lamp

Solution 2 (MCQ).

Carbon arc-lamp

3(MCQ).A radiation P is focused by a proper device on the bulb of a thermometer. Mercury in the thermometer shows a rapid increase. The radiation P is

(a) infrared radiation

(b) visible light

(c) ultraviolet radiation

(d) X-rays

Solution 3 (MCQ).

Infrared radiation
Hint: Infrared radiations produce strong heating effect.

Numericals

1.An electromagnetic wave has a frequency of 500MHz and a wavelength of 60cm.

(a) Calculate the velocity of the wave.

(b) Name the medium through which it is travelling.

Solution 1.

(a) Frequency =500MHz =500 x 106Hz
Wavelength= 60 cm=0.6 m
Velocity of wave= frequency x wavelength
= 500x 106 x 0.6=3 x 108m/s
(b) Electromagnetic wave is travelling through air.

2.The wavelength of X-rays is 0.01 Selina Solutions Concise Physics Class 10 Chapter 6 - 39. Calculate its frequency.

Solution 2.
  
Selina Concise Physics Class 10 ICSE Solutions Chapter 6 - Spectrum


Exercise 6(C)

1.What is meant by scattering of light?

Solution 1.

When white light from sun enters the earth’s atmosphere, the light gets scattered i.e., the light spreads in all directions by the dust particles, free water molecules and the molecules of the gases present in the atmosphere. This phenomenon is called scattering of light.

2.How does the intensity of scattered light depend on the wavelength of incident light? State conditions when this dependence holds.

Solution 2.

The intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light. This relation holds when the size of air molecules is much smaller than the wavelength of the light incident.

3.

Solution 3.

Violet colour is scattered the most and red the least as the intensity of scattered light is found to be inversely proportional to the fourth power of wavelength of light.

4.The danger signal is red. Why?

Solution 4.

Since the wavelength of red light is the longest in the visible light, the light of red colour is scattered the least by the air molecules of the atmosphere and therefore the light of red colour can penetrate to a longer distance. Thus red light can be seen from the farthest distance as compared to other colours of same intensity. Hence it is used for danger signal so that the signal may be visible from the far distance.

5.

Solution 5.

On the moon, since there is no atmosphere, therefore there is no scattering of sun light incident on the moon surface. Hence to an observer on the surface of moon (space), no light reaches the eye of the observer except the light directly from the sun. Thus the sky will have no colour and will appear black to an observer on the moon surface.

6.What characteristic property of light is responsible for the blue colour of the sky?

Solution 6.

Scattering property of light is responsible for the blue colour of the sky as the blue colour is scattered the most due to its short wavelength.

7.The colour of the sky, in direction of the sun is blue. Explain.

Solution 7.

As the light travels through the atmosphere, it gets scattered in different directions by the air molecules present in its path. The blue light due to its short wavelength is scattered more as compared to the red light of long wavelength. Thus the light reaching our eye directly from sun is rich in red colour, while the light reaching our eye from all other directions is the scattered blue light.

8.

Solution 8.

Therefore, the sky in direction other than in the direction of sun is seen blue.
At the time of sunrise and sunset, the light from sun has to travel the longest distance of atmosphere to reach the observer. The light travelling from the sun loses blue light of short wavelength due to scattering, while the red light of long wavelength is scattered a little, so is not lost much. Thus blue light is almost absent in sunlight reaching the observer, while it is rich in red colour.Solution 11.
At noon, the sun is above our head, so we get light rays directly from the sun without much scattering of any particular colour. Further, light has to travel less depth of atmosphere; hence the sky is seen white.

9.The clouds are seen white. Explain.

Solution 9.

The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of sizes bigger than the wavelength of visible light. Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light reaches our eye, the clouds are seen white.

1(MCQ).In the white light of sun, maximum scattering by the air molecules present in the earth’s atmosphere is for:

(a) red colour

(b) yellow colour

(c) green colour

(d) blue colour

Solution 1 (MCQ).

Blue colour
Hint: When light of certain frequency falls on that atom or molecule, this atom or molecule responds to the light, whenever the size of the atom or molecule comparable to the wavelength of light. The sizes of nitrogen and oxygen molecules in atmosphere are comparable to the wavelength of blue light. These molecules act as scattering centers for scattering of blue light. This is also the reason that we see the sky as blue.

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