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JEE Main 2026 (22 Jan, Shift 2) · Wave Theory, Wavefronts & Doppler Effect

Wave Theory, Wavefronts & Doppler Effect question from JEE Main 2026 (22 Jan, Shift 2)

Q4JEE Main 2026 · 22 JAN · SHIFT 2Easy

Wave Theory, Wavefronts & Doppler Effect

The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 4/34/3. The wavelength of the same light when it is passing through a transparent medium having refractive index of 3/23/2 is ______ nm.

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Answer: (c)

  1. λmedium=λ0/μ\lambda_{medium} = \lambda_0/\mu, so λ1μ1=λ2μ2\lambda_1\mu_1 = \lambda_2\mu_2.
  2. 540×43=λ2×32540 \times \frac{4}{3} = \lambda_2 \times \frac{3}{2}
  3. λ2=7201.5=480\lambda_2 = \dfrac{720}{1.5} = 480 nm
  4. So the answer is (c) 480 nm.
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