Question
Why don't we have interference when two candles are placed close to each other and the intensity is seen at a distant screen? What happens if the candles are replaced by laser sources?
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Answer
In order to get interference, the sources should be coherent, i.e. they should emit wave of the same frequency and a stable phase difference. Two candles that are placed close to each other are distinct and cannot be considered as coherent sources. Two independent sources cannot be coherent. So, two different laser sources will also not serve the purpose.
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$(a)\ -32\ Nm ^2 C ^{-1}$
$(b)\ -24 \ Nm ^2 C ^{-1}$
$(c)\ 32 \ Nm ^2 C ^{-1}$
$(d)\ 24 \ Nm ^2 C ^{-1}$
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$(a)\ S _2$ and $S _4$
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$OR$
The dimensional formula of surface integral $\oint \vec{E} \cdot d \vec{S}$ of an electric field is
$(a)\ \left[ M ^{-1} L^3 T^{-3} A\right]$
$(b)\ \left[ M L ^2 T^{-2} A^{-1}\right]$
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$(a)\ -24 Nm ^2 C ^{-1}$
$(b)\ 32 Nm ^2 C ^{-1}$
$(c)\ -32 Nm ^2 C ^{-1}$
$(d)\ 24 Nm ^2 C ^{-1}$
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$(a)\ -32\ Nm ^2 C ^{-1}$
$(b)\ -24 \ Nm ^2 C ^{-1}$
$(c)\ 32 \ Nm ^2 C ^{-1}$
$(d)\ 24 \ Nm ^2 C ^{-1}$
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$(a)\ S _2$ and $S _4$
$(b)\ S_3$ and $S_6$
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$(d)\ S _1$ and $S _3$
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$(a)\ 24\ Nm ^2 C ^{-1}$
$(b)\ 8\ Nm ^2 C ^{-1}$
$(c)\ -8\ Nm ^2 C ^{-1}$
$(d)$ zero
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The dimensional formula of surface integral $\oint \vec{E} \cdot d \vec{S}$ of an electric field is
$(a)\ \left[ M ^{-1} L^3 T^{-3} A\right]$
$(b)\ \left[ M L ^2 T^{-2} A^{-1}\right]$
$(c) \ \left[ ML ^3 T^{-3} A^{-1}\right]$
$(d)\ \left[ M L ^{-3} T^{-3} A^{-1}\right]$
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→|
Particle
|
Mass $\times c^2 (MeV)$
|
Kinetic energy $(MeV)$
|
|
Neutron
|
$940.97$ | $0.00$ |
|
Proton
|
$939.67$ | $0.01$ |
|
Electron
|
$0.51$ | $0.39$ |
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