Coherent light of frequency 6.37×1014 Hz passes through 2 thin slits and also drops on a display screen 88.0 cm ameans. You observe that the
The distance between the two slits is 40.11 μm.
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Distance of the screen l = 88.0 cm
Position of the 3rd order y =3.10 cm
We must calculate the wavelength
Using formula of wavelength
wright here, c = rate of light
f = frequency
Put the value right into the formula
We must calculate the distance between the two slits
Where, m = number of fringe
d = distance in between the two slits
Put the value into the formula
Hence, The distance in between the 2 slits is 40.11 μm.
A parallel combicountry of a 1.13-μF capacitor and a 2.85-μF one is associated in series to a 4.25-μF capacitor. This three-capaci
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Let C₁ , C₂ and C₃ are the capacitor which are associated to the battery having actually voltage V. According to the difficulty, C₁ and C₂ are linked in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the over equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Because, C₄ and C₃ are associated in series, tright here tantamount capacitance is:
Substitute 4.25 μF for C₃ and also 3.98 μF for C₄ in the above equation.
C₅ = 2.05 μF
The charge on the tantamount capacitance is recognize by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and also 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Because, the capacitors C₃ and also C₄ are linked in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
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Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ =
Since, C₁ and also C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.