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Physics · Geometrical Optics

JEE Main 2025 — 3 April, Evening Shift — Question 50

A monochromatic light of frequency 5×1014 Hz5 \times 10^{14} \mathrm{~Hz} travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :

  1. Option A:

    400 nm

  2. Option B:

    300 nm

    Correct
  3. Option C:

    600 nm

  4. Option D:

    500 nm

Answer: B

Step-by-step solution

v=vλv=v \lambda As μ=2\mu=2 therefore; speed c→C2c \rightarrow \frac{C}{2}

λr=C2v=3×1082×5×1014=3×108−15 m=3×10−7×109 nm=300 nm\begin{aligned} \lambda_{r} & =\frac{C}{2 v}=\frac{3 \times 10^{8}}{2 \times 5 \times 10^{14}}=3 \times 10^{8-15} \mathrm{~m} \\ & =3 \times 10^{-7} \times 10^{9} \mathrm{~nm}=300 \mathrm{~nm} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Introduction to Refraction of Light (Snell's Law)

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