Question 1

Question 6The frequency of violet light is 7.5 x 1014 Hz. Its wavelength in nm is :7500 nm4000 nm400 nm750 nm

Accepted Answer

400 nmReason — As we know,Speed of light (c) = Frequency (f) x Wavelength (λ)andC = 3 x 108 m s-1Frequency range = 7.5 × 1014 HzSubstituting the values in the formula,3×108=7.5×1014×λλ=3×1087.5×1014⇒λ=0.4×10−6m⇒λ=400×10−9m3 \times 10^8 = 7.5 × 10^{14} \times λ \[0.5em] \text {λ} = \dfrac {3 \times 10^8 }{7.5 × 10^{14}} \[0.5em] \Rightarrow \text {λ} = 0.4 \times 10^{-6} \text {m} \[0.5em] \Rightarrow \text {λ} = 400 \times 10^{-9} \text {m} \[0.5em]3×108=7.5×1014×λλ=7.5×10143×108​⇒λ=0.4×10−6m⇒λ=400×10−9mTherefore,λ = 400 nm

Question 2

Question 9Assertion (A): A beam of white light gives a spectrum on passing through a hollow prism.Reason (R): The speed of light outside the prism is same as the speed of light inside the prism.Both A and R are true and R is the correct explanation of ABoth A and R are true and R is not the correct explanation of Aassertion is false but reason is trueassertion is true but reason is false.

Accepted Answer

assertion is false but reason is trueExplanationAssertion (A) is false. A beam of white light does give a spectrum when passing through a prism, but it is not a "hollow" prism that causes this dispersion. A hollow prism contains air in it, so when a beam of white light is passed through it, it does not give a spectrum. Because light is passed to air through air, so no spectrum will be produced.Reason (R) is true. The speed of light inside a medium, such as a glass prism, is generally slower than the speed of light in vacuum but here prism is hollow so refractive indices of air and hollow is are same . So the speed of light outside the prism is same as the speed of light inside the glass prism.

Question 3

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

Accepted Answer

As we know,Speed of light (c) = Frequency (f) x Wavelength (λ)Given,λ = 550 nm= 550 × 10-9 m= 0.55 x 10-6 mC = 3 x 108 m s-1Substituting the values in the formula, we get,3×108=f×0.55×10−6f=3×1080.55×10−6⇒f=5.45×1014Hz3 \times 10^8 = \text {f} \times 0.55 × 10^{-6} \[0.5em] \text {f} = \dfrac {3 \times 10^8 }{0.55 × 10^{-6}} \[0.5em] \Rightarrow \text {f} = 5.45 \times 10^{14} \text {Hz} \[0.5em]3×108=f×0.55×10−6f=0.55×10−63×108​⇒f=5.45×1014HzHence, we get,Frequency = 5.4 × 1014 Hz

Question 4

Question 2The 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 x 108 m s-1

Accepted Answer

As we know,Speed of light (c) = Frequency (f) x Wavelength (λ)Given,C = 3 x 108 m s-1Frequency range = 3.75 × 1014 Hz to 7.5 × 1014 Hz.Let, f1 = 3.75 × 1014 HzSubstituting the values in the formula,3×108=3.75×1014×λλ=3×1083.75×1014⇒λ=0.8×10−6m⇒λ=8000×10−10m3 \times 10^8 = 3.75 × 10^{14} \times λ \[0.5em] \text {λ} = \dfrac {3 \times 10^8 }{3.75 × 10^{14}} \[0.5em] \Rightarrow \text {λ} = 0.8 \times 10^{-6} \text {m} \[0.5em] \Rightarrow \text {λ} = 8000 \times 10^{-10} \text {m} \[0.5em]3×108=3.75×1014×λλ=3.75×10143×108​⇒λ=0.8×10−6m⇒λ=8000×10−10mAs, 1 Å = 10-10 mTherefore,λ1 = 8000 ÅLet, f2 = 7.5 × 1014 HzSubstituting the values in the formula,3×108=7.5×1014×λλ=3×1087.5×1014⇒λ=0.4×10−6m⇒λ=4000×10−10m3 \times 10^8 = 7.5 × 10^{14} \times λ \[0.5em] \text {λ} = \dfrac {3 \times 10^8 }{7.5 \times 10^{14}} \[0.5em] \Rightarrow \text {λ} = 0.4 \times 10^{-6} \text {m} \[0.5em] \Rightarrow \text {λ} = 4000 \times 10^{-10} \text {m} \[0.5em]3×108=7.5×1014×λλ=7.5×10143×108​⇒λ=0.4×10−6m⇒λ=4000×10−10mAs, 1 Å = 10-10 mTherefore,λ2 = 4000 ÅTherefore, the wavelength range is 4000 Å to 8000 Å.

Question 5

Question 10How would the dispersion of light change if the prism is submerged in water instead of air ? Will the spread of the constituent colours remain the same ?

Accepted Answer

When a prism is submerged in water instead of air, the dispersion of light decreases. This is because dispersion depends on the difference in refractive index between the prism material and the surrounding medium.In air, the difference is large, so light bends more at the surfaces of the prism, causing greater dispersion. In water, the difference is smaller, so light bends less as it enters and exits the prism. As a result, the spread of the constituent colours is reduced.

Question 6

Question 11Name two properties of ultraviolet radiations which are similar to the visible light and two which differ from visible light.

Accepted Answer

Two properties of ultraviolet radiations which are similar to the visible light are as follows:Ultraviolet radiations travel in a straight line like visible light, with a speed of 3 x 108 m s-1 in air or vacuum.They obey the laws of reflection and refraction like visible light.Two properties of ultraviolet radiations which differ from the visible light are as follows:Ultraviolet radiations have wavelength in the range of 100 Å to 4000 Å whereas wavelength of visible light is the range of 4000 Å to 8000 Å.Ultraviolet radiations cause fluorescence on striking a zinc-sulphide screen whereas visible light does not cause such fluorescence.

Question 7

Question 12Mention two properties of infrared radiations which are similar to the visible light and two which differ from visible light.

Accepted Answer

Two properties of infrared radiations which are similar to the visible light are as follows:Infrared radiations travel in a straight line like visible light, with a speed of 3 x 108 m s-1 in air or vacuum.They obey the laws of reflection and refraction like visible light.Two properties of infrared radiations which differ from the visible light are as follows:Infrared radiations have wavelength in the range of 8000 Å to 107 Å whereas wavelength of visible light is the range of 4000 Å to 8000 Å.Infrared radiations do not affect ordinary photographic plate whereas visible light affects photographic plate.

Question 8

Question 13State one harmful effect each of the (a) ultraviolet and (b) infrared radiations.

Accepted Answer

Harmful effects of the given radiations are as follows —(a) Ultraviolet radiations cause health hazards like skin cancer if human body is exposed to them for a long period.(b) Infrared radiations causes skin burns.

Question 9

Question 14Give 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.

Accepted Answer

(i) Infrared radiations are used for photography in fog because they are not much scattered, and so they can penetrate appreciably through it.(ii) Infrared radiations are used for signals during the war because they are not visible and they are not absorbed much in the medium.(iii) The photographic darkrooms are provided with infrared lamps because they provide some visibility without affecting the photographic film.(iv) A rock salt prism is used instead of a glass prism to obtain the infrared spectrum because the rock salt prism does not absorb the infrared radiations, whereas a glass prism absorbs them.(v) A quartz prism is required for obtaining the spectrum of the ultraviolet light because ultraviolet radiations can pass through quartz whereas ordinary glass absorbs them.(vi) Ultraviolet bulbs have a quartz envelope instead of glass because ultraviolet radiations are not absorbed by quartz whereas ordinary glass absorbs the ultraviolet light.

Question 10

Question 15Radiations from the sun fall on a prism and suffer dispersion. Three thermometers are kept on which the radiations after dispersion are made to fall as shown in the figure given below. Which thermometer would show a higher reading ? Give a reason for your answer.

Accepted Answer

Thermometer A (with red light) would show a higher reading.Reason — Red light has the longest wavelength and the lowest frequency among the three. Hence, it carries more heat energy compared to violet or indigo light. As a result, it causes a greater rise in temperature, making thermometer A show a higher reading.

Question 11

Question 3In the white light of Sun, maximum scattering by the air molecules present in the earth’s atmosphere is for —red colouryellow colourgreen colourblue colour

Accepted Answer

blue colourReason — Intensity of scattered light is found to be inversely proportional to the fourth power of the wavelength of light.(i.e., I ∝ 1λ4\dfrac{1}{λ^4}λ41​ ).As blue light has lowest wavelength, hence, the blue light is scattered the most.

Question 12

Question 4Violet light is scattered nearly ............... times more than the red light.48169

Accepted Answer

16Reason — According to Rayleigh scattering, the amount of scattering is inversely proportional to the fourth power of the wavelength of the light.We know,λviolet = 4000 Å andλred = 8000 Å.According to Rayleigh scattering,Intensity of scattering of violetIntensity of scattering of red\dfrac{	ext{Intensity of scattering of violet}}{	ext{Intensity of scattering of red}}Intensity of scattering of redIntensity of scattering of violet​ = (λredλviolet)4(\dfrac{λ_{red}}{λ_{violet}})^4(λviolet​λred​​)4So,Intensity of scattering of violetIntensity of scattering of red\dfrac{	ext{Intensity of scattering of violet}}{	ext{Intensity of scattering of red}}Intensity of scattering of redIntensity of scattering of violet​ = (80004000)4(\dfrac{8000}{{4000}})^4(40008000​)4 = 24= 16Hence, Intensity of scattering of violet = 16 time Intensity of scattering of red.

Question 13

Question 5How would the sky appear when seen from the space (or moon)? Give reason for your answer.

Accepted Answer

When seen from the space (or moon), the sky would appear black in colour.As there is no atmosphere on moon, therefore, no scattered Sun light reaches the moon's surface. Hence, to an observer on the surface of moon, no light reaches his eyes, except the light reaching directly from the Sun.Thus, the sky in direction other than the that of the Sun will appear black.Similarly, when an astronaut goes above the atmosphere of the earth in a rocket, he sees the earth black.

Question 14

Question 7The colour of the sky, in direction other than of the Sun is blue. Explain.

Accepted Answer

The light from the Sun has to travel a long distance of the earth's atmosphere before reaching us. As light travels through the atmosphere, it gets scattered in different directions by the air molecules present in it's path.The blue (or violet) light due to its short wavelength is scattered more as compared to the red light of longer 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.Therefore, the sky in the direction, other than the direction of Sun, is seen blue.

Question 15

Question 8Why does the Sun appear red at sunrise and sunset?

Accepted Answer

At the time of sunrise and sunset, the light from Sun has to travel the longest distance of atmosphere to reach the observer.Since, the blue light of short wavelength is scattered more, much of it is lost, while the red light of long wavelength is scattered a little, so it is not much lost.Thus, blue light is almost absent in sunlight reaching the observer and only the red (white - blue = red) light reaches us.As a result, the Sun and the region near by it, is seen red.

Spectrum

Exercise 6A Multiple Choice Type

Exercise 6A Numericals

Exercise 6A Short Questions

Exercise 6B Short Question

Exercise 6C Multiple Choice Type

Exercise 6C Short Questions