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NEET PHYSICSWAVE OPTICSEasy

Question

Two periodic waves of intensities I1I_1 and I2I_2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

A

I1+I2I_1+I_2

B

(I1+I2)2(\sqrt{I_1}+\sqrt{I_2})^2

C

(I1I2)2(\sqrt{I_1}-\sqrt{I_2})^2

D

2(I1+I2)2(I_1+I_2)

Step-by-Step Solution

  1. Identify the formulas for maximum and minimum intensity: When two waves of intensities I1I_1 and I2I_2 interfere, the resultant intensity is given by I=I1+I2+2I1I2cos(Δϕ)I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos(\Delta\phi) . The maximum intensity (constructive interference) is Imax=(I1+I2)2=I1+I2+2I1I2I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = I_1 + I_2 + 2\sqrt{I_1I_2} . The minimum intensity (destructive interference) is Imin=(I1I2)2=I1+I22I1I2I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = I_1 + I_2 - 2\sqrt{I_1I_2} .
  2. Calculate the sum: Add the expressions for maximum and minimum intensities. Imax+Imin=(I1+I2+2I1I2)+(I1+I22I1I2)I_{\max} + I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2}) Imax+Imin=2I1+2I2=2(I1+I2)I_{\max} + I_{\min} = 2I_1 + 2I_2 = 2(I_1 + I_2)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSperiodicintensitiesthroughregiondirection

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