Fresnel Diffraction by a Straight Edge

Position of Maxima and Minima in Straight Edge Diffraction

Conditions: Point source, cylindrical/spherical wavefront, distance b from edge to screen, distance a from source to edge.

• Express the position coordinate y on the screen relative to the geometrical shadow boundary.
• Define the dimensionless Fresnel parameter v as $v = y \sqrt{\frac{2(a+b)}{\lambda ab}}$.
• Locate the extreme values of intensity by differentiating the Fresnel integral amplitudes, leading to condition equations involving tangents of phase parameters.
• Set the derivative of the intensity with respect to v to zero to obtain the transcendental equation for maxima and minima: tan(\frac{pi v^2}{2}) = \frac{pi v^2}{2}.

Result: The nth maximum or minimum occurs when v satisfies the condition for the turning points on Cornu's spiral, with positions $y_n = v_n \sqrt{\frac{\lambda ab}{2(a+b)}}$.

Intensity at the Geometrical Shadow Boundary

Conditions: Light passing a semi-infinite screen evaluated right at the edge shadow line (v = 0).

• Write the general expression for the complex amplitude in terms of Fresnel integrals C(v) and S(v) as $U = \frac{A_0}{\sqrt{2}} [(0.5 + C(v)) + i(0.5 + S(v))]$.
• At the geometrical shadow boundary, the upper edge of the visible wavefront corresponds to v = 0.
• Substitute v = 0 into the Fresnel integrals, knowing that C(0) = 0 and S(0) = 0.
• Calculate the intensity I as proportional to the squared modulus of the complex amplitude: I propto |U|^2 = \frac{A_0^2}{2} [(0.5)^2 + (0.5)^2] = \frac{A_0^2}{4}$.

Result: $I = \frac{I_0}{4}$, where I_0 is the uniform intensity that would be observed in the absence of the obstacle.

Intensity vs. Distance (Straight Edge Diffraction Pattern)

Draw: A graph with position y on the horizontal axis (where y=0 is the shadow edge, positive is illuminated, negative is shadow) and intensity I/I_0 on the vertical axis. Show oscillations in the positive y region decaying to 1, reaching 0.25 at y=0, and dropping monotonically to 0 in the negative y region.

Interpret: Clearly demonstrates light bending into the geometrical shadow and interference fringes outside the shadow.

Exam use: Frequently asked for sketch-based MCQs or qualitative interpretation of fringe visibility in TGT/PGT exams.

Cornu's Spiral Plot

Draw: A symmetric clothoid curve spiraling asymptotically towards the pole points P_1(0.5, 0.5) and P_2(-0.5, -0.5) in the C(v)-S(v) plane.

Interpret: Vectors drawn from P_2 to any point on the spiral represent the complex amplitude for a straight edge diffraction setup.

Exam use: Tested conceptually in PGT physics for graphical problem-solving and integral interpretations.

Optical Bench Setup for Straight Edge Diffraction

Draw: A schematic diagram showing a sodium vapor lamp (source), a narrow slit, a razor blade (straight edge) mounted vertically, and an eyepiece/micrometer screw gauge on a viewing screen bench.

Interpret: Measures the fringe spacing to experimentally verify Fresnel diffraction equations and calculate light wavelength lambda.

Exam use: Practical viva questions and arrangement-based queries in teaching exams.

A monochromatic light of wavelength 589 nm is incident on a straight edge. The distance from the source to the edge is 1 m, and the distance from the edge to the screen is 1 m. Find the value of the Fresnel parameter v for a point on the screen located at a distance of 1 mm from the geometrical shadow boundary.

• Identify given parameters: lambd$a = 589 \times 10^{-9} m$, a = 1 m, b = 1 m, $y = 1 \times 10^{-3} m$.
• Write the formula for the Fresnel parameter: $v = y \sqrt{\frac{2(a+b)}{\lambda ab}}$.
• Substitute the values: $v = (1 \times 10^{-3}) \sqrt{\frac{2(1 + 1)}{(589 \times 10^{-9}) \times (1) \times (1)}}$.
• Simplify the expression under the square root: \frac{4}{589 \times 10^{-9}} = 6.791 \times 10^6.
• Calculate the square root: \sqrt{6.791 \times 10^6} = 2606.
• Multiply by y: $v = 1 \times 10^{-3} \times 2606 = 2.606$.

Answer: 2.606

In a straight edge diffraction experiment, light of wavelength 5000 A is used with a point source at a distance of 50 cm from the edge and a screen at a distance of 50 cm from the edge. Calculate the intensity of light at the geometrical shadow edge relative to the unobstructed intensity.

• Recall the theoretical derivation for the intensity at the geometrical shadow edge (where the boundary parameter v = 0).
• At v = 0, the Fresnel integrals are C(0) = 0 and S(0) = 0.
• The complex amplitude is $U = \frac{A_0}{\sqrt{2}} [(0.5 + 0) + i(0.5 + 0)] = \frac{A_0(1 + i)}{2\sqrt{2}}$.
• The intensity I is proportional to |U|^2, so $I = |U|^2 = \frac{A_0^2}{8} \times 2 = \frac{A_0^2}{4}$.
• Since the unobstructed intensity I_0 corresponds to amplitude A_0^2, the boundary intensity is I_0 / 4.

Answer: 0.25 I_0 (or one-fourth of the unobstructed intensity)

Using the approximate formula for the first maximum in straight edge diffraction, find the distance y_1 of the first bright fringe from the edge shadow line, given a = 1 m, b = 1 m, and lambda = 6000 A.

• Convert wavelength to SI units: lambd$a = 6000 \times 10^{-10} m = 6 \times 10^{-7} m$.
• Identify distance parameters: a = 1 m, b = 1 m.
• Use the position formula for the first maximum: y_1 \approx \sqrt{\frac{\lambda ab}{a+b} \left( 1 - \frac{1}{16} \right)}.
• Substitute numerical values: \frac{\lambda ab}{a+b} = \frac{6 \times 10^{-7} \times 1 \times 1}{1 + 1} = 3 \times 10^{-7} m.
• Evaluate the bracket term: 1 - \frac{1}{16} = \frac{15}{16} = 0.9375.
• Multiply and take the square root: $y_1 = \sqrt{3 \times 10^{-7} \times 0.9375} = \sqrt{2.8125 \times 10^{-7}} = \sqrt{28.125 \times 10^{-8}} \approx 5.30 \times 10^{-4} m$.

Answer: 0.53 mm (or 5.30 \times 10^{-4} m)