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A parallel beam of light passes through a slit of width 0.45 mm. The screen is 50 cm away, where a fringe pattern is formed. The distance between two second-order minima is: (The wavelength of light is 450 nm)

a. 0.45 mm
b. 0.90 mm
c. 1.35 mm
d. 1.80 mm

User Lostaman
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1 Answer

6 votes

Final answer:

The distance between the two second-order minima in the diffraction pattern formed by a parallel beam of light passing through a slit is 0.90 mm, calculated by using the single-slit diffraction formula and trigonometry.

"the correct option is approximately option B"

Step-by-step explanation:

The question involves calculating the distance between two second-order minima in a diffraction pattern when a parallel beam of light passes through a slit and onto a screen. To compute this, we use the single-slit diffraction formula:

d sin(\(θ\)) = m\(λ\), where d is the width of the slit, \(θ\) is the angle of the minima from the central axis, m is the order of the minima, and \(λ\) is the wavelength of the light. For second-order minima (m=2), this formula can be adjusted to find the angle \(θ\) that corresponds to the minima.

Working out the angle for the minima allows us to calculate the position on the screen using simple trigonometry, considering the screen-slit distance (L). Using the formula x = L tan(\(θ\)), where x is the position of the minima on the screen, we can find the distance between the two second-order minima (x2-x1 = 2x, since they are symmetric).

The given parameters are: d = 0.45 mm, \(λ\) = 450 nm, and L = 50 cm. By substituting these into the formulas, and working through the calculations, we find that the distance between the two second-order minima corresponds to the option (b) 0.90 mm.

User Nitin Saxena
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