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  4. The distance travelled by light in glass (refractive index = 1.5) in a nanosecond will be

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Q. The distance travelled by light in glass (refractive index $= 1.5$) in a nanosecond will be

Chhattisgarh PMTChhattisgarh PMT 2004

Solution:

Speed of light in glass $v=\frac{c}{\mu}$
So, distance travelled $d=c_{g} . t =\frac{c}{\mu} \cdot t$
$=\frac{3 \times 10^{8}}{1.5} \times 1 \times 10^{-9}=0.2 \,m=20\, cm$