Answer:
None of the wavelength is in the visible range
Explanation:
Constructive interference of the reflected waves for different wavelengths can be estimated using:
λ
= 2nd/m
where m is 1,2,3, ...
Therefore:
m=1, λ
= 750 nm
m=2, λ
= 750/2 = 375 nm
The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.
Answer:
The answer to the question above is explained below
Explanation:
The reaction quotient, Q, is a measure of the relative amounts of reactants and products during a chemical reaction as it can be used to determine in which direction a reaction will proceed at a given point in time. Equilibrium constant is the numerical value of reaction quotient at the end of the reaction, when equilibrium is reached.
If Q = K then the system is already at equilibrium. If Q < Keq, the reaction will move toward the products to reach equilibrium. If Q > Keq, the reaction will move toward the reactants in order to reach equilibrium. Therefore, by comparing Q and K, we can determine the direction of a reaction.
Where Q= reaction quotient and Keq= equilibrium constant for the reaction.
The larger the equilibrium constant, the further the equilibrium lies toward the products. Reaction quotient is a quantity that changes as a reaction system approaches equilibrium.
We can determine the equilibrium constant based on equilibrium concentrations. K is the constant of a certain reaction when it is in equilibrium. Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products.
Answer:
= -1.984 Diopters
Explanation:
The lens should form an upright, virtual image at the far point.
Therefore;
The image distance, V will be -50.4 cm, since the image is virtual.
For objects at very far distant the object distance,u, will be ∞ ; This means that focal length, f, will be equivalent to image distance, v, that is -50.4 cm
Therefore; f = -50.4/100 = -0.504 m
But, since Power of a lens, P, is given by the reciprocal of focal length in meters, (1/f)
Then, power will be given by;
Power = 1/f
= 1/-0.504 m
=- 1.984
Power is measured in Diopters
Hence <u>= -1.984 Diopters</u>
Explanation:
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