Q17Engineering Physics

Question

Calculate the angles at which the first dark band and the next bright band are formed in the Fraunhofer diffraction pattern of a slit $0.3 mm$ wide ( $λ = 5890 \overset{˚}{A}$ ).

Answer 1

Fraunhofer diffraction occurs when plane wavefronts of monochromatic light (originating from a source at optical infinity) strike a narrow rectangular slit. Because of the wave nature of light, it does not cast a sharp geometric shadow. Instead, the light bends around the edges and interferes, producing a brilliant, wide central maximum bounded by alternating dark and bright bands (fringes) that gradually fade in intensity.

Mathematical Condition for Minima (Dark Fringes)

By applying Huygens' principle and integrating the path differences of secondary wavelets originating across the entire width of the slit, physics dictates that completely destructive interference (zero intensity, a dark fringe) occurs exactly when the overall path difference between wavelets from the extreme top and bottom edges of the slit is an integer multiple of the full wavelength $λ$ . This is expressed by the fundamental equation:

a sin θ = nλ

Where: - $a$ is the precise width of the slit. - $θ$ is the angular deviation from the central normal axis. - $λ$ is the wavelength of the incident light. - $n = 1, 2, 3...$ corresponds to the order of the minimum (the first dark fringe, second dark fringe, etc.). Note that $n = 0$ is strictly reserved for the Central Maximum.

Calculation of the First Order Minimum Angle

We are tasked with finding the angle $θ$ for the very first dark fringe, meaning we set $n = 1$ . First, we must ensure all physical quantities are in standard SI units (meters) to avoid devastating order-of-magnitude errors.

Slit width ( $a$ ) = $0.3 mm = 0.3 \times 1 0^{- 3} m = 3 \times 1 0^{- 4} m$
Wavelength ( $λ$ ) = $5890 \overset{˚}{A} = 5890 \times 1 0^{- 10} m = 5.89 \times 1 0^{- 7} m$
Order ( $n$ ) = 1

Rearranging the equation to solve for the sine of the angle:

sin θ_{1} = \frac{nλ}{a}

Substituting the values:

sin θ_{1} = \frac{1 \times ( 5.89 \times 1 0 ^{- 7} m )}{3 \times 1 0 ^{- 4} m}

sin θ_{1} = 1.9633 \times 1 0^{- 3} = 0.0019633

Finally, we take the inverse sine ( $arcsin$ ) to find the actual physical angle. Because the value is so incredibly small, the small-angle approximation ( $sin θ \approx θ$ in radians) is highly valid here. Using a calculator:

θ_{1} = arcsin (0.0019633) \approx 0.112 5^{\circ}

This incredibly tight angle (roughly one-tenth of a single degree) demonstrates why diffraction effects are generally unnoticeable in everyday macroscopic life, becoming prominent only when the aperture is microscopically small.

Q17Engineering Physics

Question

Answer

Mathematical Condition for Minima (Dark Fringes)

Calculation of the First Order Minimum Angle