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Optics - Interference of Light (Young's Double Slit)

Grade 12ICSEPhysics

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

🔑Concepts

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Interference is the phenomenon of redistribution of light energy in a medium due to the superposition of light waves from two coherent sources.

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Coherent sources are sources of light that emit waves of the same frequency (or wavelength) and have a constant phase difference ϕ\phi between them.

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The Principle of Superposition states that the resultant displacement y⃗\vec{y} at any point is the vector sum of individual displacements: y⃗=y1⃗+y2⃗\vec{y} = \vec{y_1} + \vec{y_2}.

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Constructive Interference: Occurs at points where the waves meet in the same phase. The path difference is an integral multiple of wavelength: Δx=nλ\Delta x = n\lambda, where n=0,1,2,…n = 0, 1, 2, \dots.

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Destructive Interference: Occurs at points where the waves meet in opposite phase. The path difference is an odd multiple of half-wavelength: Δx=(2n−1)λ2\Delta x = (2n - 1)\frac{\lambda}{2}, where n=1,2,3,…n = 1, 2, 3, \dots.

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Fringe Width (β\beta): The distance between two consecutive bright fringes or two consecutive dark fringes. In YDSE, the bright and dark fringes are of equal width.

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The intensity of light II is proportional to the square of the amplitude a2a^2. Resultant intensity is given by I=I1+I2+2I1I2cos⁡ϕI = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi.

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Conditions for sustained interference: Sources must be coherent, monochromatic, and the distance between slits dd must be small compared to the distance to the screen DD.

📐Formulae

Δx=dsin⁡θ≈xdD\Delta x = d \sin \theta \approx \frac{xd}{D}

xn=nλDd (Position of nth bright fringe)x_n = \frac{n \lambda D}{d} \text{ (Position of } n^{th} \text{ bright fringe)}

xn=(2n−1)λD2d (Position of nth dark fringe)x_n = (2n - 1) \frac{\lambda D}{2d} \text{ (Position of } n^{th} \text{ dark fringe)}

β=λDd\beta = \frac{\lambda D}{d}

θ=λd (Angular fringe width)\theta = \frac{\lambda}{d} \text{ (Angular fringe width)}

Imax=(a1+a2)2 and Imin=(a1−a2)2I_{max} = (a_1 + a_2)^2 \text{ and } I_{min} = (a_1 - a_2)^2

ImaxImin=(r+1r−1)2 where r=a1a2=I1I2\frac{I_{max}}{I_{min}} = \left( \frac{r+1}{r-1} \right)^2 \text{ where } r = \frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}}

💡Examples

Problem 1:

In a Young's double slit experiment, the slits are separated by 0.28 mm0.28 \text{ mm} and the screen is placed 1.4 m1.4 \text{ m} away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm1.2 \text{ cm}. Determine the wavelength of light used.

Solution:

Given: d=0.28 mm=0.28×10−3 md = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}, D=1.4 mD = 1.4 \text{ m}, n=4n = 4, and x4=1.2 cm=1.2×10−2 mx_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}. Using the formula xn=nλDdx_n = \frac{n \lambda D}{d}, we rearrange for λ\lambda: λ=xndnD\lambda = \frac{x_n d}{n D}. Substituting the values: λ=(1.2×10−2)(0.28×10−3)4×1.4=0.336×10−55.6=6×10−7 m=600 nm\lambda = \frac{(1.2 \times 10^{-2}) (0.28 \times 10^{-3})}{4 \times 1.4} = \frac{0.336 \times 10^{-5}}{5.6} = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}.

Explanation:

The position of the nthn^{th} bright fringe from the center is given by xn=nβx_n = n\beta. By substituting the distance of the 4th fringe, we can calculate the wavelength λ\lambda.

Problem 2:

Find the ratio of intensities at two points PP and QQ on a screen in YDSE, where waves from two sources have path differences of 00 and λ4\frac{\lambda}{4} respectively.

Solution:

Phase difference ϕ=2πλΔx\phi = \frac{2\pi}{\lambda} \Delta x. For point PP, Δx=0  ⟹  ϕP=0\Delta x = 0 \implies \phi_P = 0. For point QQ, Δx=λ4  ⟹  ϕQ=2πλ⋅λ4=π2\Delta x = \frac{\lambda}{4} \implies \phi_Q = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}. Using I=Imaxcos⁡2(ϕ2)I = I_{max} \cos^2(\frac{\phi}{2}), we have IP=Imaxcos⁡2(0)=ImaxI_P = I_{max} \cos^2(0) = I_{max}. For QQ, IQ=Imaxcos⁡2(π4)=Imax(12)2=Imax2I_Q = I_{max} \cos^2(\frac{\pi}{4}) = I_{max} (\frac{1}{\sqrt{2}})^2 = \frac{I_{max}}{2}. Thus, IPIQ=ImaxImax/2=2:1\frac{I_P}{I_Q} = \frac{I_{max}}{I_{max}/2} = 2:1.

Explanation:

Intensity in interference patterns depends on the phase difference ϕ\phi. We convert path difference to phase difference and then use the cosine squared intensity distribution formula.

Interference of Light (Young's Double Slit) Revision - Class 12 Physics ICSE