Class 12 Physics Wave Optics Validity of wave optics

Validity of wave optics

  • Consider a single slit of an aperture of thickness ‘a’.Diffraction pattern will be observed.
  • There will be centralmaxima due to diffraction.
  • Angular size of central maximum = (λ/a)
    • where θ (condition to have central maximum)=(λ/a)
  • The distance travelled by the light waves for the spread to occur = ‘z’(the angular spread due to diffraction)
    • The spread is given as = (z λ/a)
    • Where (z λ/a) = a where a =width of the diffracted beam.
  • z ≈ (a2/ λ).This distance is known as Fresnel distance.
  • Fresnel distance describes the distance at which spread due to diffraction becomes comparable to the width of the slit or not.
  • This is the boundary of ray optics and wave optics.
  • For distances << zf :
    • Spreading due to diffraction is smaller compared to size of the beam then the ray optics is valid.
  • For distances >> zf :
  • Spreading due to the diffraction dominates over ray optics then the wave optics is valid.

Problem: Estimate the distance for which ray optics is good approximation for an aperture of 4 mmand wavelength 400 nm.

Answer:- Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. Itis given by the relation,

ZF = (a2/λ)


Aperture width, a = 4 mm = 4 ×10−3 m

Wavelength of light, λ = 400 nm = 400 × 10−9 m

ZF = (4x10-3)2/ (400x10-9)

Therefore, the distance for which the ray optics is a good approximation is 40 m.

Problem:- A slit 4cm wide is irradiated with microwaves of wavelength 2cm.Find the angular spread of central maximum, assuming incidence normal to the plane of the slit?

Answer:- Distance between the slits d=4cm,λ=2cm

The half angular speed of central maxima is given by,

d sinθ1 =1 x λ =λ

Therefore sinθ1 =(λ/d) = (2/4) =0.5

Or θ1 =300.

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