Listed below are alterations made, one at a time, to the original experiment, and the experiment is repeated. after each alterat
ion, the experiment is returned to its original configuration. which of these alterations decreases the angles at which the diffraction minima appear
1 answer:
Answer : The question is incomplete, full question is attached in the answer.
The equation that will be applicable to given situation will be as;
d X sinΘ = m X λ;
<span>Here, d will be the slit width,
</span>Θ will be the angle of the minima,
m will be the order of the minima,
and λ will be the wavelength of the light.
Now we have to find Θ, On rearranging the equation we get it as; Θ = arc sin( m X λ<span> / d )
In the given condition of A when the d increase the argument increases with the angle of </span>Θ;
In case of <span>B it seems to be</span> irrelevant, as changing the distance to the screen will not affect the distance between the diffraction angle of minima, it will only change its width;
In case of C the angle of Θ decreases, as we decrease the argument the d value also decreases;
In D it compensates to decrease the λ and hence the Θ also decreases;
In E when it is observed i<span>n water, the light's frequency remains to be the same, but it speed of the light ray slows down, so the wavelength gets smaller. So, it will decrease the dispersion angle of minima;
</span>
In F it seems to irrelevant as it will not change the angle of dispersion.
The equation that will be applicable to given situation will be as;
d X sinΘ = m X λ;
<span>Here, d will be the slit width,
</span>Θ will be the angle of the minima,
m will be the order of the minima,
and λ will be the wavelength of the light.
Now we have to find Θ, On rearranging the equation we get it as; Θ = arc sin( m X λ<span> / d )
In the given condition of A when the d increase the argument increases with the angle of </span>Θ;
In case of <span>B it seems to be</span> irrelevant, as changing the distance to the screen will not affect the distance between the diffraction angle of minima, it will only change its width;
In case of C the angle of Θ decreases, as we decrease the argument the d value also decreases;
In D it compensates to decrease the λ and hence the Θ also decreases;
In E when it is observed i<span>n water, the light's frequency remains to be the same, but it speed of the light ray slows down, so the wavelength gets smaller. So, it will decrease the dispersion angle of minima;
</span>
In F it seems to irrelevant as it will not change the angle of dispersion.
You might be interested in
Answer: .
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
(1)
(2)
The final reaction is:
(3)
By subtracting (1) and (2)
Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ
Freezing point is the temperature at which a substance undergoes a phase transition from liquid to solid state. Water converts from liquid state to solid ice at a temperature of .
Mass is the amount of matter contained in a substance. The SI unit of mass is kilogram. 1 kilogram is equal to 1000 g. One kilogram is defined as the mass of one cubic decimeter of water at the freezing point of water.