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CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

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Question : 7 of 13
Marks: +1, -0
A ray of light travels a distance of 12.0 m12.0\,\mathrm{m} in a transparent sheet in 60 ns60\,\mathrm{ns}. The refractive index of the sheet is
Solution:     👈: Video Solution
Speed of light in the sheet   =  12 m60×10−9 s\;=\;\frac{12\,\mathrm{m}}{60 \times 10^{-9}\,\mathrm{s}}
  =2×108 ms−1\;=2 \times 10^8\,\mathrm{ms}^{-1}
Speed of light in vacuum =3×108 ms−1=3 \times 10^8\,\mathrm{ms}^{-1}
So, the refractive index of the sheet =  3×1082×108=1.5=\;\frac{3 \times 10^8}{2 \times 10^8}=1.5
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