Because each fault line has different types of pressure being held, and some fault lines can hold more pressure before being released, resulting a larger eq, and some other lines hold less pressure resulting in a smaller eq... hope I was helpful :)
Answer:
(a). The thickness of the glass is 868 nm.
(b). The wavelength is 3472 nm.
Explanation:
Given that,
Refractive index = 1.20
Wavelength = 496 nm
Next wavelength = 386 nm
We need to calculate the thickness of the glass
Using formula for constructive interference
Put the value into the formula
In first case,
.....(I)
In second case,
.....(II)
From equation (I) and (II)
Put the value of m in equation (I)
The thickness of the glass is 868 nm.
(b). We need to calculate the wavelength
Using formula of constructive interference
Put the value into the formula
Hence, (a). The thickness of the glass is 868 nm.
(b). The wavelength is 3472 nm.
Heyyyy I wish I could help but I don’t know
Answer:
(1) 1×10⁻⁴
Explanation:
From the question,
α = (ΔL/L)/(ΔT)............. Equation 1
Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.
Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C
Substitute these values into equation 1
α = 1×10⁻⁴/10
α = 1×10⁻⁵ °C⁻¹ .
β = (ΔA/A)/ΔT................... Equation 2
Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.
make ΔA/A the subject of the equation
ΔA/A = β×ΔT.......................... Equation 3
But,
β = 2α.......................... Equation 4
Substitute equation 4 into equation 3
ΔA/A = 2α×ΔT................ Equation 5
Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹
Substitute into equation 5
ΔA/A = ( 2)×(1×10⁻⁵)×(5)
ΔA/A = 10×10⁻⁵
ΔA/A = 1×10⁻⁴
Hence the right option is (1) 1×10⁻⁴
Answer:
When a ray of light passes through a glass slab of a certain thickness, the ray gets displaced or shifted from the original path. This is called lateral shift/displacement.
Explanation:
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