Answer:
(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.
Note: find an attached copy of a part of the solution to the given question below.
Explanation:
Solution
Recall that:
d = 6.6 um
λ₀ =d/10
λ₀ = 6.6 um
Now,
(a) We find the wavelength λ of the light in water.
Thus,
λ water = (λ₀ )/n
= 0.66/1.33
So,
λ water = λ = 0.496 um
(b) We find the phase difference between the waves from slit 1 and 2
Now,
if a <<d and a<<λ
Then the path difference between the rays will be
Δ S₂N = Δ d sinθ
Thus, the phase difference becomes,
S = 2π Δ/λ is S= 2π Δ d sinθ/ λ
<u>(</u>c) The next step is to derive an expression for the intensity I as function of O and other relevant parameters.
Now,
Let p be the point where these two rays interfere with each other.
Thus,
The electric field vector coming out from slot and and slot 2 is
E₁= E₀₁ cos (ks₁ p - wt) i
E₂ = E₀₂ cos (ks₂ p - wt) i
Note: Kindly find an attached copy of a part of the solution to the given question below.
If the object's <em>velocity is constant</em> ... (it's speed isn't changing AND it's moving in a straight line) ... then the net force on the object is zero.<em> (D)</em>
Either there are no forces at all acting on the object, OR there are forces on it but they're 'balanced' ... when you add up all of their sizes and directions, they just exactly cancel each other out, and they have the SAME EFFECT on the object as if there were no forces at all.
Answer:
Explanation:
Given,
- Work done by the rope 900 m/s.
- Angle of inclination of the slope =
- Initial speed of the skier = v = 1.0 m/s
- Length of the inclined surface = d = 8.0 m
part (a)
The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity
In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.
part (b)
- Initial speed of the skier = v = 1.0 m/s.
Rate of the work done by the rope is power of the rope.
Part (c)
- Initial speed of the skier = v = 2.0 m/s.
Rate of the work done by the rope is power of the rope.
