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State the reason, why two independent sources of light cannot be considered as coherent sources.
Two independent sources of light cannot be considered coherent because coherence requires a fixed phase relationship between the waves. Independent sources generally emit light with random phases, causing interference patterns to fluctuate over time. This lack of consistent phase correlation means tRead more
Two independent sources of light cannot be considered coherent because coherence requires a fixed phase relationship between the waves. Independent sources generally emit light with random phases, causing interference patterns to fluctuate over time. This lack of consistent phase correlation means they are not spatially or temporally coherent, essential for producing stable interference effects.
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What are coherent sources of light?
Coherent sources of light emit waves with a constant phase difference, the same frequency, and wavelength. These sources maintain a stable phase relationship over time, enabling the formation of clear and stable interference patterns. They are essential for experiments like Young’s double-slit. ForRead more
Coherent sources of light emit waves with a constant phase difference, the same frequency, and wavelength. These sources maintain a stable phase relationship over time, enabling the formation of clear and stable interference patterns. They are essential for experiments like Young’s double-slit.
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How does the fringe width in Young’s double-slit experiment change when the distance of separation ‘D’ between the slits and screen is doubled?
In Young’s double-slit experiment, fringe width (w) is directly proportional to the distance (D) between the slits and the screen, given by w = λD/d . When D is doubled, the fringe width also doubles. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-11/
In Young’s double-slit experiment, fringe width (w) is directly proportional to the distance (D) between the slits and the screen, given by w = λD/d . When D is doubled, the fringe width also doubles.
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Define work function for a given metallic surface.
The work function of a metallic surface is the minimum energy required to eject an electron from the metal's surface. It depends on the material and is typically measured in electron volts (eV). For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-11/
The work function of a metallic surface is the minimum energy required to eject an electron from the metal’s surface. It depends on the material and is typically measured in electron volts (eV).
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A bullet is fired and gets embedded in a block kept on table. If table is frictionless, then
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation. 1. Momentum Conservation: - During the collision, the system or the bullet along with the block is isolated; no external force acts on the systemRead more
In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation.
1. Momentum Conservation:
– During the collision, the system or the bullet along with the block is isolated; no external force acts on the system in the horizontal direction (since the table is frictionless). Hence, momentum is conserved.
– Before the collision, the block is at rest, and the bullet is moving with its momentum. After the bullet gets embedded in the block, the total momentum is conserved before the collision to be equal to the total momentum after the collision.
2. Kinetic Energy Conservation:
– In this inelastic collision, kinetic energy is not conserved. Some of the kinetic energy from the bullet goes into internal energy during the collision (such as heat, sound, and deformation of the bullet and block).
3. Potential Energy Conservation:
– Potential energy is not involved in this problem since it does not change due to the fact that the bullet is embedded into the block horizontally on a frictionless table.
Final Answer:
In this case, momentum is conserved.
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