krit.club logo

Wave Optics - Interference and Young’s Double Slit Experiment

Grade 12CBSEPhysics

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

🔑Concepts

•

Superposition Principle: When two or more wave pulses overlap, the resultant displacement at any point is the vector sum of the displacements of the individual waves: y⃗=y1⃗+y2⃗\vec{y} = \vec{y_1} + \vec{y_2}.

•

Coherent Sources: Two sources are said to be coherent if they emit light waves of the same frequency and have a constant phase difference ϕ\phi. Coherent sources are necessary for a stable interference pattern.

•

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

•

Constructive Interference (Maxima): Occurs when waves meet in phase. The resultant intensity is maximum. Path difference Δx=nλ\Delta x = n\lambda and phase difference ϕ=2nπ\phi = 2n\pi, where n=0,1,2,...n = 0, 1, 2, ...

•

Destructive Interference (Minima): Occurs when waves meet out of phase. The resultant intensity is minimum. Path difference Δx=(2n−1)λ2\Delta x = (2n-1)\frac{\lambda}{2} and phase difference ϕ=(2n−1)π\phi = (2n-1)\pi, where n=1,2,...n = 1, 2, ...

•

Young's Double Slit Experiment (YDSE): A setup using two narrow slits S1S_1 and S2S_2 separated by distance dd to produce interference fringes on a screen at distance DD.

•

Fringe Width (β\beta): The distance between two consecutive bright fringes or two consecutive dark fringes. It is directly proportional to wavelength λ\lambda and screen distance DD, and inversely proportional to slit separation dd.

•

Intensity Distribution: In YDSE, all bright fringes have the same intensity Imax=4I0I_{max} = 4I_0 (if I1=I2=I0I_1 = I_2 = I_0), and dark fringes have zero intensity (for perfectly coherent, equal amplitude sources).

📐Formulae

Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta x

I=I1+I2+2I1I2cos⁡ϕI = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi

Imax=(I1+I2)2 and Imin=(I1−I2)2I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \text{ and } I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2

yn(bright)=nλDd for n=0,1,2...y_n(\text{bright}) = \frac{n\lambda D}{d} \text{ for } n = 0, 1, 2...

yn(dark)=(2n−1)λD2d for n=1,2,3...y_n(\text{dark}) = (2n-1)\frac{\lambda D}{2d} \text{ for } n = 1, 2, 3...

β=λDd\beta = \frac{\lambda D}{d}

Angular Fringe Width θ=βD=λd\text{Angular Fringe Width } \theta = \frac{\beta}{D} = \frac{\lambda}{d}

💡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 (n=4n=4) bright fringe is measured to be 1.2 cm1.2\text{ cm}. Determine the wavelength of light used in the experiment.

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 y4=1.2 cm=1.2×10−2 my_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}. Using the formula for the position of the nthn^{th} bright fringe: yn=nλDdy_n = \frac{n\lambda D}{d}. Rearranging for λ\lambda: λ=yndnD\lambda = \frac{y_n d}{nD}. Substituting values: λ=(1.2×10−2 m)×(0.28×10−3 m)4×1.4 m=6×10−7 m=600 nm\lambda = \frac{(1.2 \times 10^{-2} \text{ m}) \times (0.28 \times 10^{-3} \text{ m})}{4 \times 1.4 \text{ m}} = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}.

Explanation:

The position of the nthn^{th} bright fringe relative to the central maximum is directly proportional to the order nn, wavelength λ\lambda, and distance DD, and inversely proportional to slit width dd.

Problem 2:

The ratio of intensities of two waves in an interference pattern is 25:925:9. Calculate the ratio of maximum to minimum intensity (Imax/IminI_{max}/I_{min}) in the resulting pattern.

Solution:

Given I1I2=259\frac{I_1}{I_2} = \frac{25}{9}. Since intensity I∝a2I \propto a^2, the ratio of amplitudes is a1a2=I1I2=53\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \frac{5}{3}. Let a1=5ka_1 = 5k and a2=3ka_2 = 3k. The ratio ImaxImin=(a1+a2)2(a1−a2)2=(5k+3k)2(5k−3k)2=(8k)2(2k)2=64k24k2=16\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{(5k + 3k)^2}{(5k - 3k)^2} = \frac{(8k)^2}{(2k)^2} = \frac{64k^2}{4k^2} = 16. Thus, Imax:Imin=16:1I_{max}:I_{min} = 16:1.

Explanation:

Maximum intensity occurs when amplitudes add constructively (a1+a2a_1 + a_2), and minimum intensity occurs when they interfere destructively (a1−a2a_1 - a_2). The intensity is the square of the resultant amplitude.

Interference and Young’s Double Slit Experiment Revision - Class 12 Physics CBSE