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If the distance between the first maxima and fifth minima of a double slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength (in nm) of the light used is :
    Correct answer is '600'. Can you explain this answer?
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    If the distance between the first maxima and fifth minima of a double ...
    There are three and a half fringes from first maxima to fifth minima as shown

    The correct answer is: 600
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    If the distance between the first maxima and fifth minima of a double ...
    To determine the wavelength of light used in a double-slit experiment, we can use the equation:

    d * sinθ = m * λ

    where d is the distance between the slits, θ is the angle of the maxima or minima, m is the order of the maxima or minima, and λ is the wavelength.

    In this case, we are given the distance between the first maxima and fifth minima as 7 mm. To find the wavelength, we need to find the value of sinθ for both the maxima and minima.

    Let's break down the problem into steps:

    Step 1: Find the value of sinθ for the first maxima.
    - The distance between the first maxima and the central bright fringe (zeroth order) is given by:
    d * sinθ = λ
    - Since the first maxima is at a distance of 7 mm from the central bright fringe, we have:
    d * sinθ = 7 mm
    - Plugging in the values:
    0.15 mm * sinθ = 7 mm
    - Simplifying:
    sinθ = 7 mm / 0.15 mm
    sinθ = 46.67

    Step 2: Find the value of sinθ for the fifth minima.
    - The distance between the central bright fringe (zeroth order) and the fifth minima is also given by:
    d * sinθ = (2m + 1) * λ
    - Since the fifth minima is at a distance of 7 mm from the central bright fringe, we have:
    d * sinθ = (2 * 5 + 1) * λ
    d * sinθ = 11λ
    - Plugging in the values:
    0.15 mm * sinθ = 11λ
    - Simplifying:
    sinθ = 11λ / 0.15 mm
    sinθ = 73.33λ

    Step 3: Find the wavelength.
    - Now, we can equate the values of sinθ for the first maxima and fifth minima:
    46.67 = 73.33λ
    - Solving for λ:
    λ = 46.67 / 73.33
    λ ≈ 0.636 nm

    Therefore, the wavelength of the light used in the double-slit experiment is approximately 0.636 nm, which can be rounded to 600 nm.
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    If the distance between the first maxima and fifth minima of a double slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength (in nm) of the light used is :Correct answer is '600'. Can you explain this answer?
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