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 ≈ (a**^{2}/**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 << z
_{f}:- Spreading due to diffraction is smaller compared to size of the beam then the ray optics is valid.

- For distances >> z
_{f}: - 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 (Z

Z_{F} = (a^{2}/λ)

Where,

Aperture width, a = 4 mm = 4 ×10^{−3} m

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

Z_{F} = (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 sinθ_{1} =1 x λ =λ

Therefore sinθ_{1} =(λ/d) = (2/4) =0.5

Or θ_{1} =30^{0}.

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