(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10.0º, at what angle is the second-order maximum?
(b) What is the angle of the first minimum?
(c) What is the highest-order maximum possible here?
(a)y(max) = 2k Lλ/2d = k Lλ/d
tan φ= k Lλ/L d = k λ/d
tan φ₁=1•λ/d
tanφ₂=2•λ/d
tan φ₂=2•tan φ₁ 2•tan10
φ₂ ≃20⁰
(b) y(min) = (2k+1) Lλ/2d
tan φ = (2k+1) Lλ/2Ld =(2k+1) λ/2d
tan φ₋₁=λ/2d = tan φ₁/2 =tan10/2
φ₋₁ ≃5⁰
(c)tan φ₁=1•λ/d
Max wavelength of visual range id λ= 0.76•10⁻⁶ m λ(max)
d= λ/tan φ₁= 0.76•10⁻⁶/tan10=4.3•10⁻⁶ m
k=d sinφ/ λ
sinφ≤1 . Take sinφ=1
k= 4.3•10⁻⁶/0.76•10⁻⁶=5.67.
k(max) = 5
(a) To determine the angle of the second-order maximum, we can use the formula for the double-slit interference pattern, which is given as:
sinθ = mλ/d
Where:
θ is the angle of the maximum or minimum,
m is the order of the maximum,
λ is the wavelength of the light, and
d is the distance between the two slits.
From the given information, we know that the first-order maximum occurs at an angle of 10.0º, so m = 1. We need to find the angle for the second-order maximum, so m = 2.
Using the formula, we have:
sinθ2 = 2λ/d
Since we're looking for the angle, we can rearrange the equation to solve for θ2:
θ2 = arcsin(2λ/d)
Please provide the values for λ and d to calculate the angle of the second-order maximum.
(b) To find the angle of the first minimum, we know that the first minimum occurs when the path difference between the two waves is λ/2. The formula for the angle of the first minimum is given as:
sinθ_min = (λ/2)/d
Rearranging the equation to solve for θ_min:
θ_min = arcsin((λ/2)/d)
Please provide the values for λ and d to calculate the angle of the first minimum.
(c) The highest-order maximum possible can be determined by considering the conditions for constructive interference. The formula for the highest-order maximum is given as:
sinθ_max = (m + 1/2)λ/d
To determine the highest-order maximum, we need to know the values of λ and d.
Please provide the values for λ and d to calculate the highest-order maximum.
To answer these questions, we need to use the concept of diffraction of light. The angle at which the maxima and minima occur in a double-slit experiment can be determined using the formula:
dsinθ = mλ,
where d is the slit separation, θ is the angle of the maximum or minimum, m is the order of the maximum or minimum, and λ is the wavelength of light.
(a) To find the angle of the second-order maximum, we can use the formula mentioned above. We know that the first-order maximum occurs at θ = 10.0º. Let's assume the slit separation, d, and the wavelength, λ, remain constant. We can rewrite the formula as:
dsinθ = mλ.
Substituting the values, we have:
d x sin(10.0º) = 1 x λ.
Now, to find the angle of the second-order maximum, we need to find the value of θ when m = 2. Rearranging the formula, we have:
sinθ = (2 x λ) / d.
Using the inverse sine function (sin^-1), we can find the angle:
θ = sin^-1((2 x λ) / d).
(b) To find the angle of the first minimum, we can use the same formula. The first minimum occurs when m = 1, so we have:
sinθ = (1 x λ) / d.
Again, using the inverse sine function, we can find the angle:
θ = sin^-1((1 x λ) / d).
(c) The highest-order maximum possible can be found when the value of m is maximized. Since m represents the order of the maximum, it can theoretically go to infinity. However, in practice, the intensity of the maxima decreases significantly as the order increases, making higher-order maxima difficult to observe. In most cases, only a few orders of maxima are visible, and the higher-order maxima are too faint to be perceived.
Hence, the highest-order maximum possible depends on the experimental setup and the observer's ability to detect faint intensity patterns. Generally, the first few orders (m = 0, 1, 2, etc.) are considered significant and observable.