Skip to content
Question 192 of 199

Q.Obtain the relation between phase difference and path difference.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2025Subjective· 3mImportance★★★★★
96% · 192/199 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Superposition and Interference of Light

Interference is the redistribution of light intensity that results when two light waves overlap (are superposed) at a point, producing bright regions (increased intensity) at some points and dark regions (decreased intensity) at others, in a way that a single wave alone never would. If two waves of the same frequency and amplitudes a1,a2a_1,a_2 arrive at a point with a constant phase difference ϕ\phi between them, y1=a1sin⁡ωty_1=a_1\sin\omega t and y2=a2sin⁡(ωt+ϕ)y_2=a_2\sin(\omega t+\phi), their resultant is y=Asin⁡(ωt+θ)y=A\sin(\omega t+\theta) with resultant amplitude A2=a12+a22+2a1a2cos⁡ϕA^2=a_1^2+a_2^2+2a_1a_2\cos\phi; since intensity I∝A2I\propto A^2, the resultant intensity is I∝I1+I2+2I1I2cos⁡ϕI\propto I_1+I_2+2\sqrt{I_1I_2}\cos\phi. Constructive interference (maximum brightness) occurs when ϕ=0,±2π,±4π,…\phi=0,\pm2\pi,\pm4\pi,\ldots, giving Imax⁡∝(a1+a2)2∝I1+I2+2I1I2I_{\max}\propto(a_1+a_2)^2\propto I_1+I_2+2\sqrt{I_1I_2}; destructive interference (minimum brightness, possibly complete darkness) occurs when ϕ=±π,±3π,…\phi=\pm\pi,\pm3\pi,\ldots, giving Imin⁡∝(a1−a2)2∝I1+I2−2I1I2I_{\min}\propto(a_1-a_2)^2\propto I_1+I_2-2\sqrt{I_1I_2}. For the special case of equal amplitudes (a1=a2=aa_1=a_2=a, I1=I2=I0I_1=I_2=I_0), this simplifies …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.