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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.