Answer:
Explanation:
The problem is based on interference in thin films
refractive index of water is more than given oil so there will be phase change of π at upper and lower layer of the film .
a )
for constructive interference , the condition is
2μt = nλ where t is thickness of layer , μ is refractive index , λ is wavelength and n is order of the fringe
Putting the values
2 x 1.27 t = n x 640
2 x 1.27 t = 640 ( for minimum thickness n = 1 )
t = 252 nm .
b )
2 x 1.27 t = m₁ λ₁
for destructive interference
2μt = (2m₂+1)λ₂/2
2 x 1.27 t =(2m₂+1)λ₂/2
m₁ λ₁ = (2m₂+1)λ₂/2
2m₁λ₁ = (2m₂+1)λ₂
2m₁ / (2m₂+1) = λ₂ / λ₁
2m₁ / (2m₂+1) = 548/ 640
2m₁ / (2m₂+1) = .85625
2m₁ = .85625 (2m₂+1)
This is the required relation between m₁ and m₂
Answer:
1832
Explanation:
From;
Δp Δx = h/4π
Δp = uncertainty in momentum
Δx = uncertainty in position
h= Plank's constant
But p =mv hence, Δp= Δmv
m= mass, v= velocity
mass of electron = 9.11 * 10^-31 Kg
Mass of proton = 1.67 * 10^-27 Kg
since m is a constant,
Δv = h/Δxm4π
For proton;
Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10
Δv = 315 ms-1
For electron;
Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10
Δv = 577000 ms-1
Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832
This shows the sunlight shining on the moon is creating a shadow. The part of the shadow that we don’t see from Earth is called the eclipse. Since the Earth gets a shadow from the moon, that causes the rest of the Earth to light up which causes a lunar eclipse.
Explanation:
The correct answer would be C. three half lives.
1. Current means flow of electrons and electrons are negative charged and so are attracted to the positive end of the battery and repelled by the negative end.
2. When battery is connected in a circuit that lets the electron flow through it, they flow from negative to positive.
3. When negative terminal of cell is connected to other negative terminal of the cell in a particular circuit then, current will not flow in circuit as electrons cannot flow from negative to negative terminal.
