Angle of minimum deviation, δm = 40° and angle of the prism, A = 60° Refractive index of water, μ = 1.33 and refractive index of the material of the prism =μ' The angle of deviation is related to refractive index (μ') as: μ' = [sin (A+δm)/2]/sin A/2 = [sin (60°+40°)/2]/sin 60°/2 = [sin (50°)]/sin 30Read more
Angle of minimum deviation, δm = 40° and angle of the prism, A = 60°
Refractive index of water, μ = 1.33 and refractive index of the material of the prism =μ’
The angle of deviation is related to refractive index (μ’) as:
μ’ = [sin (A+δm)/2]/sin A/2
= [sin (60°+40°)/2]/sin 60°/2
= [sin (50°)]/sin 30°= 1.532
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let δ’m be the new angle of minimum deviation for the same prism. The refractive index of glass with respect to water is given by the relation:
μgw = μ’/μ = [sin (A+δ’m)/2]/sin A/2
=> [sin (A+δ’m)/2] = μ’/μ x (sin A/2)
=> [sin (A+δ’m)/2] = 1.532/1.33 x sin 60°/2= 0.5759
=> (A+δ’m)/2 = sin⁻10.5759=35.16°
60° +δ’m = 70.32°
Therefore ,
δ’m = 70.32° – 60°
= 10.32°
Hence, the new minimum angle of deviation is 10.32°.
As per the given figure, for the glass - air interface: Angle of incidence, i = 60° Angle of refraction, r = 35° The relative refractive index of glass with respect to air is given by Snell's law as: μga= sin i/sin r = sin 60°/sin 35°= 0.8660/0.5736 = 1.51 --------------Eq-1 As per the given figure,Read more
As per the given figure, for the glass – air interface:
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as:
μga= sin i/sin r
= sin 60°/sin 35°= 0.8660/0.5736 = 1.51 ————–Eq-1
As per the given figure, for the air-water interface:
Angle of incidence, i = 60°
Angle of refraction r= 47°
The relative refractive index of water with respect to air is given by Snell’s law as :
μwa= sin i/sin r
= sin 60°/sin 47°= 0.8660/0.7314 = 1.184————–Eq-2
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
μgw = μga/μwa
=1.51/1.184= 1.275
The following figure shows the situation involving the glass – water interface.
Angle of incidence, i = 45°
Angle of refraction = r
From Snell’s law, r can be calculated as:
sin i/sin r= μgw
=>sin 45°/sin r = 1.275
sin r = (1/√2)/1.275 = 0.5546
Therefore ,r = sin⁻¹ (0.5546)
= 38.68°
Hence, the angle of refraction at the water – glass interface is 38.68°.
Actual depth of the needle in water, hi = 12.5 cm and apparent depth of the needle in water, h2 = 9.4 cm Refractive index of water = μ The value of μ can be obtained as follows: μ = h₁/h₂ = 12.5/9.4 ≈ 1.33 Hence, the refractive index of water is about 1.33. When water is replaced by a liquid of refrRead more
Actual depth of the needle in water, hi = 12.5 cm and apparent depth of the needle in water, h2 = 9.4 cm
Refractive index of water = μ
The value of μ can be obtained as follows:
μ = h₁/h₂ = 12.5/9.4 ≈ 1.33
Hence, the refractive index of water is about 1.33.
When water is replaced by a liquid of refractive index, μ’ = 1.63
The actual depth of the needle remains the same, but its apparent depth changes.
Let y be the new apparent depth of the needle. Hence, we can write the relation:
μ’= h₁/y
=> y = h₁/μ’ = 12.5/1.63 = 7.67 cm
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.
Therefore, distance by which the microscope should be moved up = 9.4 – 7.67 = 1.73 cm
Height of the needle, h1 = 4.5 cm Object distance, u = -12 cm Focal length of the convex mirror, f = 15 cm Image distance = v The value of v can be obtained using the mirror formula: 1/u + 1/v = 1/f 1/v = 1/f -1/u 1/v = 1/15 - (1/-12) = 1/15 + 1/12 = (4+5 )/60 = 9/60 Therefore , v = 60/9 = 6.7 cm HeRead more
Height of the needle, h1 = 4.5 cm
Object distance, u = -12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
1/u + 1/v = 1/f
1/v = 1/f -1/u
1/v = 1/15 – (1/-12)
= 1/15 + 1/12 = (4+5 )/60 = 9/60
Therefore , v = 60/9 = 6.7 cm
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
m = h2/h1= -v/u
Therefore , h2 = -v/u x h1
=(-6.7/-12) x 4.5 = + 2.5 cm
Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
Size of the candle, h = 2.5 cm Image size = h' Object distance, u = -27 cm Radius of curvature of the concave mirror, R = -36 cm Focal length of the concave mirror, f =R /2 = -18 cm Image distance = v The image distance can be obtained using the mirror formula: 1/f= 1/u + 1/v Read more
Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
Focal length of the concave mirror, f =R /2 = -18 cm
Image distance = v
The image distance can be obtained using the mirror formula:
1/f= 1/u + 1/v “
=> 1/v = 1/f -1/u = 1/-18 – 1/-27= -1/54
=> v = —54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
m’ = h’/h = -v/u
Therefore , h’ = -v/u x h
= – (-54/-27) x 2.5 = -5 cm
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
Ans (a). Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere. Ans (b). It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not refRead more
Ans (a).
Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
Ans (b).
It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
Ans (c).
With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth, (d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
Ans (d).
In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
Ans (e).
A global nuclear war on the surface of the Earth would have disastrous consequences.
Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
Ans (a). Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum. Ans (b). Radio waves; it belongs to the short wavelength end. Ans (c). Temperature, T = 2.7 °K , λmis given by Planck's law as: λm = 0.29/2.7= 0.11 cm This wavelength corresponds to microwaves. Ans (d). ThiRead more
Ans (a).
Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
Ans (b).
Radio waves; it belongs to the short wavelength end.
Ans (c).
Temperature, T = 2.7 °K , λmis given by Planck’s law as:
λm = 0.29/2.7= 0.11 cm
This wavelength corresponds to microwaves.
Ans (d).
This is the yellow light of the visible spectrum.
Ans (e).
Transition energy is given by the relation,
E = hν, Where, h = Planck’s constant = 6.6 x 10-34 Js,
A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck's law. It can be given by the relation, λm= 0.29/T cm K Where, λm= maximum wavelength and T = temperaRead more
A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
λm= 0.29/T cm K
Where,
λm= maximum wavelength and T = temperature.
Thus, the temperature for different wavelengths can be obtained as:
For λm = 10⁻4 cm; T = 0.29/10⁻4 = 2900 °K
For λm = 5 x 10⁻5 cm; T = 0.29/(5 x 10⁻5) = 5800 °K
For λm = 10⁻⁶ cm; T = 0.29/( 10⁻⁶) = 290000 °K and so on.
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
Power rating of bulb, P = 100 W It is given that about 5% of its power is converted into visible radiation. Therefore ,Power of visible radiation, P' = (5/100 ) x 100 = 5 W Hence, the power of visible radiation is 5W. Ans (a). Distance of a point from the bulb, d = 1 m Hence, intensity of radiationRead more
Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
Therefore ,Power of visible radiation,
P’ = (5/100 ) x 100 = 5 W
Hence, the power of visible radiation is 5W.
Ans (a).
Distance of a point from the bulb, d = 1 m Hence, intensity of radiation at that point is given as:
I = P’/(4πd²) =5/4π(1)² =0.398 W/m²
Ans (b).
Distance of a point from the bulb, d1 = 10 m Hence, intensity of radiation at that point is given as:
Ans (a). From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e. —J. Ans (b). It is given that, E = 3.1 N/C cos[(l.8 rad/m)y+(5.4 x 10⁸ rad/s)t]i --------------ERead more
Ans (a).
From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e. —J.
Ans (b).
It is given that,
E = 3.1 N/C cos[(l.8 rad/m)y+(5.4 x 10⁸ rad/s)t]i ————–Eq-1
The general equation for the electric field vector in the positive x direction can be written as:
E= E0 sin (kx -ωt) i—————————————-Eq-2
On comparing equations (1) and (2), we get Electric field amplitude, Eo = 3.1 N/C Angular frequency, ω = 5.4 x 108 rad/s
Wave number, k = 1.8 rad/m
Wavelength, λ = 2π/1.8 = 3.490 m
Ans (c).
Frequency of wave is given as:
ν = ω /2π = (5.4 x 108)/2π =8.6 x 10⁷ Hz
Ans (d).
Magnetic field strength is given as:
B0= E0/c
Where, c = Speed of light = 3 x 108 m/s
Therefore, B0= 3.1/(3 x 108 )= 1.03 x 10⁻⁷T
Ans (e).
On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
B= B0 cos (ky + ωt)k
= {(1.03 x10⁻7 T)cos[(1.8 rad/m)y+(5.4 x 106 rad/s)t ]}k
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Angle of minimum deviation, δm = 40° and angle of the prism, A = 60° Refractive index of water, μ = 1.33 and refractive index of the material of the prism =μ' The angle of deviation is related to refractive index (μ') as: μ' = [sin (A+δm)/2]/sin A/2 = [sin (60°+40°)/2]/sin 60°/2 = [sin (50°)]/sin 30Read more
Angle of minimum deviation, δm = 40° and angle of the prism, A = 60°
Refractive index of water, μ = 1.33 and refractive index of the material of the prism =μ’
The angle of deviation is related to refractive index (μ’) as:
μ’ = [sin (A+δm)/2]/sin A/2
= [sin (60°+40°)/2]/sin 60°/2
= [sin (50°)]/sin 30°= 1.532
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let δ’m be the new angle of minimum deviation for the same prism. The refractive index of glass with respect to water is given by the relation:
μgw = μ’/μ = [sin (A+δ’m)/2]/sin A/2
=> [sin (A+δ’m)/2] = μ’/μ x (sin A/2)
=> [sin (A+δ’m)/2] = 1.532/1.33 x sin 60°/2= 0.5759
=> (A+δ’m)/2 = sin⁻10.5759=35.16°
60° +δ’m = 70.32°
Therefore ,
δ’m = 70.32° – 60°
= 10.32°
Hence, the new minimum angle of deviation is 10.32°.
See lessFigures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45s with the normal to a water- glass interface [Fig. 9.34(c)].
As per the given figure, for the glass - air interface: Angle of incidence, i = 60° Angle of refraction, r = 35° The relative refractive index of glass with respect to air is given by Snell's law as: μga= sin i/sin r = sin 60°/sin 35°= 0.8660/0.5736 = 1.51 --------------Eq-1 As per the given figure,Read more
As per the given figure, for the glass – air interface:
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as:
μga= sin i/sin r
= sin 60°/sin 35°= 0.8660/0.5736 = 1.51 ————–Eq-1
As per the given figure, for the air-water interface:
Angle of incidence, i = 60°
Angle of refraction r= 47°
The relative refractive index of water with respect to air is given by Snell’s law as :
μwa= sin i/sin r
= sin 60°/sin 47°= 0.8660/0.7314 = 1.184————–Eq-2
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
μgw = μga/μwa
=1.51/1.184= 1.275
The following figure shows the situation involving the glass – water interface.
Angle of incidence, i = 45°
Angle of refraction = r
From Snell’s law, r can be calculated as:
sin i/sin r= μgw
=>sin 45°/sin r = 1.275
sin r = (1/√2)/1.275 = 0.5546
Therefore ,r = sin⁻¹ (0.5546)
= 38.68°
Hence, the angle of refraction at the water – glass interface is 38.68°.
See lessA tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Actual depth of the needle in water, hi = 12.5 cm and apparent depth of the needle in water, h2 = 9.4 cm Refractive index of water = μ The value of μ can be obtained as follows: μ = h₁/h₂ = 12.5/9.4 ≈ 1.33 Hence, the refractive index of water is about 1.33. When water is replaced by a liquid of refrRead more
Actual depth of the needle in water, hi = 12.5 cm and apparent depth of the needle in water, h2 = 9.4 cm
Refractive index of water = μ
The value of μ can be obtained as follows:
μ = h₁/h₂ = 12.5/9.4 ≈ 1.33
Hence, the refractive index of water is about 1.33.
When water is replaced by a liquid of refractive index, μ’ = 1.63
The actual depth of the needle remains the same, but its apparent depth changes.
Let y be the new apparent depth of the needle. Hence, we can write the relation:
μ’= h₁/y
=> y = h₁/μ’ = 12.5/1.63 = 7.67 cm
Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.
Therefore, distance by which the microscope should be moved up = 9.4 – 7.67 = 1.73 cm
See lessA 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Height of the needle, h1 = 4.5 cm Object distance, u = -12 cm Focal length of the convex mirror, f = 15 cm Image distance = v The value of v can be obtained using the mirror formula: 1/u + 1/v = 1/f 1/v = 1/f -1/u 1/v = 1/15 - (1/-12) = 1/15 + 1/12 = (4+5 )/60 = 9/60 Therefore , v = 60/9 = 6.7 cm HeRead more
Height of the needle, h1 = 4.5 cm
Object distance, u = -12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
1/u + 1/v = 1/f
1/v = 1/f -1/u
1/v = 1/15 – (1/-12)
= 1/15 + 1/12 = (4+5 )/60 = 9/60
Therefore , v = 60/9 = 6.7 cm
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
m = h2/h1= -v/u
Therefore , h2 = -v/u x h1
=(-6.7/-12) x 4.5 = + 2.5 cm
Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
See lessA small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Size of the candle, h = 2.5 cm Image size = h' Object distance, u = -27 cm Radius of curvature of the concave mirror, R = -36 cm Focal length of the concave mirror, f =R /2 = -18 cm Image distance = v The image distance can be obtained using the mirror formula: 1/f= 1/u + 1/v Read more
Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
Focal length of the concave mirror, f =R /2 = -18 cm
Image distance = v
The image distance can be obtained using the mirror formula:
1/f= 1/u + 1/v “
=> 1/v = 1/f -1/u = 1/-18 – 1/-27= -1/54
=> v = —54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
m’ = h’/h = -v/u
Therefore , h’ = -v/u x h
= – (-54/-27) x 2.5 = -5 cm
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
See lessAnswer the following questions: (a) Long distance radio broadcasts use short-wave bands. Why? (b) It is necessary to use satellites for long distance TV transmission. Why? (c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? (e) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?
Ans (a). Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere. Ans (b). It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not refRead more
Ans (a).
Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
Ans (b).
It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
Ans (c).
With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth, (d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
Ans (d).
In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
Ans (e).
A global nuclear war on the surface of the Earth would have disastrous consequences.
Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
See lessGiven below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs. (a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space). (b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift). (c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe]. (d) 5890 A º – 5896 A º [double lines of sodium] (e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)].
Ans (a). Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum. Ans (b). Radio waves; it belongs to the short wavelength end. Ans (c). Temperature, T = 2.7 °K , λmis given by Planck's law as: λm = 0.29/2.7= 0.11 cm This wavelength corresponds to microwaves. Ans (d). ThiRead more
Ans (a).
Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
Ans (b).
Radio waves; it belongs to the short wavelength end.
Ans (c).
Temperature, T = 2.7 °K , λmis given by Planck’s law as:
λm = 0.29/2.7= 0.11 cm
This wavelength corresponds to microwaves.
Ans (d).
This is the yellow light of the visible spectrum.
Ans (e).
Transition energy is given by the relation,
E = hν, Where, h = Planck’s constant = 6.6 x 10-34 Js,
ν = Frequency of radiation
Energy, E = 14.4 K eV
Therefore , ν = E/h
=(14.4 x 10³ x 1.6 x 10⁻¹⁹)/( 6.6 x 10-34 )
=3.4 x 10¹⁸Hz
This corresponds to X-rays.
See lessUse the formula λm T= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck's law. It can be given by the relation, λm= 0.29/T cm K Where, λm= maximum wavelength and T = temperaRead more
A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
λm= 0.29/T cm K
Where,
λm= maximum wavelength and T = temperature.
Thus, the temperature for different wavelengths can be obtained as:
For λm = 10⁻4 cm; T = 0.29/10⁻4 = 2900 °K
For λm = 5 x 10⁻5 cm; T = 0.29/(5 x 10⁻5) = 5800 °K
For λm = 10⁻⁶ cm; T = 0.29/( 10⁻⁶) = 290000 °K and so on.
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
See lessAbout 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.
Power rating of bulb, P = 100 W It is given that about 5% of its power is converted into visible radiation. Therefore ,Power of visible radiation, P' = (5/100 ) x 100 = 5 W Hence, the power of visible radiation is 5W. Ans (a). Distance of a point from the bulb, d = 1 m Hence, intensity of radiationRead more
Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
Therefore ,Power of visible radiation,
P’ = (5/100 ) x 100 = 5 W
Hence, the power of visible radiation is 5W.
Ans (a).
Distance of a point from the bulb, d = 1 m Hence, intensity of radiation at that point is given as:
I = P’/(4πd²) =5/4π(1)² =0.398 W/m²
Ans (b).
Distance of a point from the bulb, d1 = 10 m Hence, intensity of radiation at that point is given as:
I = P’/[4π(d1)²]
=5/4π(10)² =0.00398 W/m²
See lessSuppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1N/C) cos [(1.8 rad/m)y + (5.4 x 10⁶ rad/s)t]} l. (a) What is the direction of propagation? (b) What is the wavelength λ? (c) What is the frequency ν? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave.
Ans (a). From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e. —J. Ans (b). It is given that, E = 3.1 N/C cos[(l.8 rad/m)y+(5.4 x 10⁸ rad/s)t]i --------------ERead more
Ans (a).
From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e. —J.
Ans (b).
It is given that,
E = 3.1 N/C cos[(l.8 rad/m)y+(5.4 x 10⁸ rad/s)t]i ————–Eq-1
The general equation for the electric field vector in the positive x direction can be written as:
E= E0 sin (kx -ωt) i—————————————-Eq-2
On comparing equations (1) and (2), we get Electric field amplitude, Eo = 3.1 N/C Angular frequency, ω = 5.4 x 108 rad/s
Wave number, k = 1.8 rad/m
Wavelength, λ = 2π/1.8 = 3.490 m
Ans (c).
Frequency of wave is given as:
ν = ω /2π = (5.4 x 108)/2π =8.6 x 10⁷ Hz
Ans (d).
Magnetic field strength is given as:
B0= E0/c
Where, c = Speed of light = 3 x 108 m/s
Therefore, B0= 3.1/(3 x 108 )= 1.03 x 10⁻⁷T
Ans (e).
On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
B= B0 cos (ky + ωt)k
= {(1.03 x10⁻7 T)cos[(1.8 rad/m)y+(5.4 x 106 rad/s)t ]}k
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