When the light ray illustrated in this Figure passes through the glass block of the index of refraction $n = 1.50$ , it is shifted laterally by the distance $d$ .
Find the time interval required for the light to pass through the glass block.

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Solution

The speed of light in the material is
$v = \frac{c}{n} = \frac{3.00 \times 10^{8} m / s}{1.50} = 2.00 \times 10^{8} m / s$

The distance $h$ traveled by the light is $h = 2.12 c m .$ The time interval is
$Δ t = \frac{h}{v} = \frac{2 \times 10^{- 2} m}{2.00 \times 10^{8} m / s} = 1 \times 10^{- 10} s = 1 n s$

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When the light ray illustrated in this Figure passes through the glass block of the index of refraction n=1.50, it is shifted laterally by the distance d. Find the time interval required for the light to pass through the glass block.

The speed of light in the material isv=cn=3.00×108m/s1.50=2.00×108m/sThe distance h traveled by the light is h=2.12cm. The time interval isΔt=hv=2×10−2m2.00×108m/s=1×10−10s=1 ns

When the light ray illustrated in this Figure passes through the glass block of the index of refraction $n = 1.50$ , it is shifted laterally by the distance $d$ .
Find the time interval required for the light to pass through the glass block.