In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What
1 answer:
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =
λ₂ = 357.3 nm
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Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between and
Answer:
= 85.89 ° C
Explanation:
The linear thermal expansion process is given by
ΔL = L α ΔT
For the three-dimensional case, the expression takes the form
ΔV = V β ΔT
Let's apply this equation to our case
ΔV / V = -0.507% = -0.507 10-2
ΔT = (ΔV / V) 1 /β
ΔT = -0.507 10⁻² 1 / 1.15 10⁻³
ΔT = -4.409
–T₀ = 4,409
= T₀ - 4,409
= 90.3-4409
= 85.89 ° C