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Wave Optics - Diffraction

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Diffraction is the phenomenon of bending of light around the corners of an obstacle or aperture and its spreading into the region of the geometrical shadow. This occurs when the size of the obstacle/aperture is comparable to the wavelength λ\lambda of light.

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Fraunhofer Diffraction (Single Slit): In this class of diffraction, the source of light and the screen are effectively at infinite distances from the diffracting aperture. This is achieved using converging lenses.

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Central Maximum: The central point on the screen has a path difference of 00 and hence has the maximum intensity. Its width is double the width of any secondary maximum.

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Conditions for Minima: For a slit of width aa, the nthn^{th} minimum occurs at an angle θ\theta such that asin⁡θ=nλa \sin \theta = n\lambda, where n=1,2,3,…n = 1, 2, 3, \dots.

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Conditions for Secondary Maxima: These occur at positions where asin⁡θ=(n+12)λa \sin \theta = (n + \frac{1}{2})\lambda for n=1,2,3,…n = 1, 2, 3, \dots. The intensity decreases rapidly as nn increases.

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Fresnel Distance (ZFZ_F): It is the distance from the slit to the screen up to which the ray optics approximation is valid. It is given by ZF=a2λZ_F = \frac{a^2}{\lambda}.

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Resolving Power: The ability of an optical instrument to show two close objects as separate. For a telescope, it is limited by diffraction and is inversely proportional to the limit of resolution Δθ\Delta \theta.

📐Formulae

asin⁡θ=nλ (Condition for nth minimum)a \sin \theta = n\lambda \text{ (Condition for } n^{th} \text{ minimum)}

asin⁡θ=(n+12)λ (Condition for nth secondary maximum)a \sin \theta = (n + \frac{1}{2})\lambda \text{ (Condition for } n^{th} \text{ secondary maximum)}

θ0=λa (Angular position of first minimum)\theta_0 = \frac{\lambda}{a} \text{ (Angular position of first minimum)}

2θ=2λa (Angular width of central maximum)2\theta = \frac{2\lambda}{a} \text{ (Angular width of central maximum)}

W=2Dλa (Linear width of central maximum on screen at distance D)W = \frac{2D\lambda}{a} \text{ (Linear width of central maximum on screen at distance } D)

Δθ=1.22λD (Limit of resolution of a telescope)\Delta \theta = \frac{1.22 \lambda}{D} \text{ (Limit of resolution of a telescope)}

R.P.=1Δθ=D1.22λ (Resolving power of a telescope)R.P. = \frac{1}{\Delta \theta} = \frac{D}{1.22 \lambda} \text{ (Resolving power of a telescope)}

dmin=1.22λ2nsin⁡β (Limit of resolution of a microscope)d_{min} = \frac{1.22 \lambda}{2n \sin \beta} \text{ (Limit of resolution of a microscope)}

R.P.=2nsin⁡β1.22λ (Resolving power of a microscope)R.P. = \frac{2n \sin \beta}{1.22 \lambda} \text{ (Resolving power of a microscope)}

💡Examples

Problem 1:

A parallel beam of light of wavelength 600 nm600 \text{ nm} is incident normally on a slit of width 0.2 mm0.2 \text{ mm}. Find the angular width of the central maximum.

Solution:

Given: λ=600×10−9 m\lambda = 600 \times 10^{-9} \text{ m}, a=0.2×10−3 ma = 0.2 \times 10^{-3} \text{ m}. The angular width of the central maximum is given by 2θ=2λa2\theta = \frac{2\lambda}{a}. Substituting values: 2θ=2×600×10−90.2×10−3=6×10−3 rad2\theta = \frac{2 \times 600 \times 10^{-9}}{0.2 \times 10^{-3}} = 6 \times 10^{-3} \text{ rad}.

Explanation:

The angular width is the angle between the first minima on either side of the central maximum. Each minimum is at an angle θ=λa\theta = \frac{\lambda}{a}.

Problem 2:

Light of wavelength 500 nm500 \text{ nm} falls on a single slit of width 1μm1 \mu\text{m}. At what angle will the first maximum (secondary) be observed?

Solution:

For the first secondary maximum (n=1n=1), the condition is asin⁡θ=32λa \sin \theta = \frac{3}{2}\lambda. Given a=1×10−6 ma = 1 \times 10^{-6} \text{ m} and λ=500×10−9 m\lambda = 500 \times 10^{-9} \text{ m}, we have sin⁡θ=3×500×10−92×1×10−6=0.75\sin \theta = \frac{3 \times 500 \times 10^{-9}}{2 \times 1 \times 10^{-6}} = 0.75. Therefore, θ=sin⁡−1(0.75)≈48.6∘\theta = \sin^{-1}(0.75) \approx 48.6^\circ.

Explanation:

The secondary maxima are located approximately halfway between the minima. The first secondary maximum specifically corresponds to the path difference of 3λ2\frac{3\lambda}{2}.

Problem 3:

Determine the resolving power of a telescope whose objective lens has a diameter of 5.08 m5.08 \text{ m} and the wavelength of light used is 600 nm600 \text{ nm}.

Solution:

Given D=5.08 mD = 5.08 \text{ m} and λ=600×10−9 m\lambda = 600 \times 10^{-9} \text{ m}. Resolving Power R.P.=D1.22λR.P. = \frac{D}{1.22 \lambda}. R.P.=5.081.22×600×10−9≈6.94×106R.P. = \frac{5.08}{1.22 \times 600 \times 10^{-9}} \approx 6.94 \times 10^6.

Explanation:

Resolving power depends directly on the diameter of the objective lens; a larger aperture allows for better resolution of distant objects.

Diffraction - Revision Notes & Key Formulas | CBSE Class 12 Physics