Answer:
Yes, it will produce a diffraction pattern.
a. 3.9 mm b. 1.95 mm
Explanation:
The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.
Given
wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m
width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m
Distance of slit from central maximum , D = 1.65 m
Distance of first minimum from central maximum, y = ?
a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...
The relationship between y and D is given by
Since θ is small, sinθ ≈ θ ≈ tanθ
so, dθ = mλ ⇒ θ = mλ/d = y/D
Therefore, y = mλD/d
Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d
Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm
b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have
λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm
Answer:
option A
Explanation:
given,
compression for lower setting = x
work done to compress in lower setting = 5 J
compression in higher setting, x' = 2 x
work done in higher setting = ?
Work done in compression of spring at lower setting
............(1)
Work done in compression of spring at higher setting
from equation (1)
W' = 20 J
Work take for the higher setting is equal to 20 J
Hence, the correct answer is option A
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s