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A ray of light strikes a parallel-sided glass slab at point A, undergoing partial reflection and refraction. At each reflection, 25% of the incident energy is reflected (the rest is transmitted). The two emerging rays AB and A'B' interfere. Find the ratio of maximum to minimum intensity in the resulting interference pattern.

1. 49: 1
2. 7: 1
3. 4: 1
4. 16: 1

Correct answer: 49: 1

Solution

Let incident intensity be I. Ray AB is reflected once at the top surface: I1 = 0.25 I. Ray A'B' refracts in (0.75), reflects at the bottom surface (0.25), then refracts out (0.75): I2 = 0.75 * 0.25 * 0.75 I = 0.140625 I. Then sqrt(I1) = sqrt(0.25) = 0.5, sqrt(I2) = sqrt(0.140625) = 0.375. Imax/Imin = (0.5+0.375)²/(0.5-0.375)² = (0.875)²/(0.125)² = (0.875/0.125)² = 7² = 49. So 49: 1.

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