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CBSE Class 10 Science 2015 Term II Outside Delhi Set 1

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Question : 4 of 36
Marks: +1, -0
The absolute refractive indices of glass and water are 43\frac{4}{3} and 32\frac{3}{2} respectively. If the speed of light in glass is 2×108 m/s2 \times 10^8 \text{ m/s}, calculate the speed of light in (i) vacuum, (ii) water.
Solution:  
(i) Given: ng=43,  nw=32,  vg=2×108 m/sn_g=\frac{4}{3},\; n_w=\frac{3}{2},\; v_g=2 \times 10^8 \text{ m/s}
We know that for glass,
ng=cvgn_g=\frac{c}{v_g}
∴\therefore Speed of light in air or vacuum,
c=ngvgc = n_g v_g
  =  43×2×108=2.67×108 m/s\; = \; \frac{4}{3} \times 2 \times 10^8 = 2.67 \times 10^8 \text{ m/s}
(ii) Also in water,
nw=cvwn_w = \frac{c}{v_w}
∴    vw=cnw=2.67×2×1083\therefore \;\; v_w = \frac{c}{n_w} = \frac{2.67 \times 2 \times 10^8}{3}
Speed of light in water =1.78×108 m/s= 1.78 \times 10^8 \text{ m/s}.
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