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The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then
(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-
(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-
- a)Power radiated by the filament is in the range 642 W to 645 W
- b)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 W
- c)The wavelength corresponding to the maximum intensity of light is 1160 nm
- d)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011
Correct answer is option 'B,C,D'. Can you explain this answer?
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The filament of a light bulb has surface area 64 mm2. The filament can...
Power radiated by the filament:
The power radiated by a black body can be calculated using the Stefan-Boltzmann law:
\[P = A\epsilon\sigma T^4\]
Where P is the power, A is the surface area, ε is the emissivity (assumed to be 1 for a black body), σ is the Stefan-Boltzmann constant, and T is the temperature.
Substitute the given values:
\[P = 64 \times 10^{-6} \times 5.67 \times 10^{-8} \times 2500^4\]
\[P = 642 \, \text{W}\]
Radiated power entering into one eye of the observer:
The power entering the pupil of the observer can be calculated by considering the solid angle subtended by the filament at the observer's eye.
\[P_{\text{eye}} = \frac{P}{4\pi d^2} \times \Omega\]
Where d is the distance between the observer and the filament and Ω is the solid angle.
Substitute the given values:
\[P_{\text{eye}} = \frac{642}{4\pi \times 100^2} \times \frac{\pi r^2}{d^2}\]
\[P_{\text{eye}} = 3.2 \times 10^{-8} \, \text{W}\]
The wavelength corresponding to the maximum intensity of light:
The wavelength corresponding to the maximum intensity can be calculated using Wien's displacement law:
\[\lambda_{\text{max}} = \frac{b}{T}\]
Where b is Wien's displacement constant and T is the temperature.
Substitute the given values:
\[\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}}{2500} = 1160 \, \text{nm}\]
Total number of photons entering per second into one eye:
The total number of photons can be calculated using the energy of one photon and the power entering the eye.
\[E = hf = \frac{hc}{\lambda}\]
\[N = \frac{P_{\text{eye}}}{E}\]
Substitute the given values:
\[N = \frac{3.2 \times 10^{-8}}{\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1740 \times 10^{-9}}}\]
\[N = 2.8 \times 10^{11} \text{photons/s}\]
The power radiated by a black body can be calculated using the Stefan-Boltzmann law:
\[P = A\epsilon\sigma T^4\]
Where P is the power, A is the surface area, ε is the emissivity (assumed to be 1 for a black body), σ is the Stefan-Boltzmann constant, and T is the temperature.
Substitute the given values:
\[P = 64 \times 10^{-6} \times 5.67 \times 10^{-8} \times 2500^4\]
\[P = 642 \, \text{W}\]
Radiated power entering into one eye of the observer:
The power entering the pupil of the observer can be calculated by considering the solid angle subtended by the filament at the observer's eye.
\[P_{\text{eye}} = \frac{P}{4\pi d^2} \times \Omega\]
Where d is the distance between the observer and the filament and Ω is the solid angle.
Substitute the given values:
\[P_{\text{eye}} = \frac{642}{4\pi \times 100^2} \times \frac{\pi r^2}{d^2}\]
\[P_{\text{eye}} = 3.2 \times 10^{-8} \, \text{W}\]
The wavelength corresponding to the maximum intensity of light:
The wavelength corresponding to the maximum intensity can be calculated using Wien's displacement law:
\[\lambda_{\text{max}} = \frac{b}{T}\]
Where b is Wien's displacement constant and T is the temperature.
Substitute the given values:
\[\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}}{2500} = 1160 \, \text{nm}\]
Total number of photons entering per second into one eye:
The total number of photons can be calculated using the energy of one photon and the power entering the eye.
\[E = hf = \frac{hc}{\lambda}\]
\[N = \frac{P_{\text{eye}}}{E}\]
Substitute the given values:
\[N = \frac{3.2 \times 10^{-8}}{\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1740 \times 10^{-9}}}\]
\[N = 2.8 \times 10^{11} \text{photons/s}\]
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The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer?
Question Description
The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer?.
The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer?.
Solutions for The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer?, a detailed solution for The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? has been provided alongside types of The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then(Take Stefan-Boltzmann constant = 5.67 × 10−8 Wm−2 K−4, Wien’s displacement constant = 2.90 × 10−3 m-K, Planck’s constant = 6.63 × 10−34 Js, speed of light in vacuum= 3.00 × 108 ms−1)-a)Power radiated by the filament is in the range 642 W to 645 Wb)Radiated power entering into one eye of the observer is in the range 3.15 × 10−8 W to 3.25 × 10−8 Wc)The wavelength corresponding to the maximum intensity of light is 1160 nmd)Taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons entering per second into one eye of the observer is in the range 2.75 × 1011 to 2.85 × 1011Correct answer is option 'B,C,D'. Can you explain this answer? tests, examples and also practice JEE tests.
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