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Ask QuestionPosted by Mansi Sharma 6 years, 11 months ago
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Posted by Mansi Sharma 6 years, 11 months ago
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Yogita Ingle 6 years, 11 months ago
According to de-Broglie hypothesis, a stationary orbit is the one that contains an integral number of de-Broglie waves associated with the revolving electron.
Total distance covered by electron = Circumference of the orbit =
For the permissible orbit,
2πrn = n{tex}\lambda{/tex}........... (i)
Now according to de-Broglie wavelegnth
{tex}\lambda{/tex}= h/mv
Now puiting this in (i)
2πrn = n{tex}h/mv{/tex}.
Posted by Shashwat Kothari 6 years, 11 months ago
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Yo Yo Honey Singh ? 6 years, 11 months ago
Posted by Rasheed Thekkarayil 6 years, 11 months ago
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Sridhar Sathiyaseelan 6 years, 11 months ago
Polarisation of transverse waves:
Let a rope AB be passed through two parallel vertical slits S1 and S2 placed close to each other. The rope is fixed at the end B. If the free end A of the rope is moved up and down perpendicular to its length, transverse waves are generated with vibrations parallel to the slit. These waves pass through both S1 and S2 without any change in their amplitude. But if S2 is made horizontal, the two slits are perpendicular to each other. Now, no vibrations will pass through S2 and amplitude of vibrations will become zero. i.e the portion S2B is without wave motion.
On the otherhand, if longitudinal waves are generated in the rope by moving the rope along forward and backward, the vibrations will pass through S1 and S2 irrespective of their positions.
This implies that the orientation of the slits has no effect on the propagation of the longitudinal waves, but the propagation of the transverse waves, is affected if the slits are not parallel to each other.
A similar phenomenon has been observed in light, when light passes through a tourmaline crystal.
Light from the source is allowed to fall on a tourmaline crystal which is cut parallel to its optic axis. The emergent light will be slightly coloured due to natural colour of the crystal. When the crystal A is rotated, there is no change in the intensity of the emergent light. Place another crystal B parallel to A in the path of the light. When both the crystals are rotated together, so that their axes are parallel, the intensity of light coming out of B does not change. When the crystal B alone is rotated, the intensity of the emergent light from B gradually decreases. When the axis of B is at right angles to the axis of A, no light emerges from B.
If the crystal B is further rotated, the intensity of the light coming out of B gradually increases and is maximum again when their axis are parallel.
Comparing these observations with the mechanical analogue discussed earlier, it is concluded that the light waves are transverse in nature.
Light waves coming out of tourmaline crystal A have their vibrations in only one direction, perpendicular to the direction of propagation. These waves are said to be polarised. Since the vibrations are restricted to only one plane parallel to the axis of the crystal, the light is said to be plane polarised. The phenomenon of restricting the vibrations into a particular plane is known as polarisation.
Yogita Ingle 6 years, 11 months ago
(i) Using the phenomenon of polarisation, show hon transverse nature of light can be demonstrated.
(ii) Two polaroids P1 and P2 are placed with their pass-axes perpendicular to each other. Unpolarised light of intensity I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its pass-axis makes an angle of 30° with that of P1. Determine the intensity of light transmitted through P1, P2 and P3.
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Yogita Ingle 6 years, 11 months ago
According to Gauss’s law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The total electric flux through a closed surface is zero if no charge is enclosed by the surface.
- Gauss’s law is true for any closed surface, no matter what its shape or size.
- The term q on the right side of Gauss’s lawincludes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
- In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, represents only the total charge inside S.
- The surface that we choose for the application of Gauss’s law is called the Gaussian surface. The Gaussian surface can pass through a continuous charge distribution.
- Gauss’s law is useful for the calculation of the electrostatic field for a symmetric system.
- Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.
Posted by Varun Upreti 6 years, 11 months ago
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