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CBSE Class 10 Science 2015 Term II Delhi Set 1

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Question : 22 of 36
Marks: +1, -0
(a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum.
(b) The absolute refractive indices of two media 'A' and 'B' are 2.0 and 1.5 respectively. If the speed of light in medium ' BB ' is 2×108 m/s2 \times 10^8 \text{ m/s}, calculate the speed of light in :
(i) vacuum,
(ii) medium 'A'.
Solution:  
(a) Laws of refraction of light:
(i) Incident ray travelling from one medium to another, bend in such a way that the incident ray, refracted ray and the normal ray, at the point of incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence (i) to the sine of angle of refraction (r)(r) is constant for a given pair of media.
i.e.   sinisinr=\;\frac{\sin i}{\sin r} = constant
Absolute refractive index of a medium is the ratio of   sini\; \sin i to   sinr\; \sin r.
Absolute refractive index
=    Speed of light in vacuum    Speed of light in medium  = \;\frac{\;\text{Speed of light in vacuum}\;}{\;\text{Speed of light in medium}\;}
(b) Given: nA=2.0,nB=1.5n_{A}=2.0, n_{B}=1.5, speed of light in medium B(vB)=2×108 m/sB (v_B) = 2 \times 10^8 \text{ m/s}, speed of light in medium vA=v_{A}= ?
(i) nB  =  c  Speed of light in a medium  (vB)n_{B} \;=\; \frac{c}{\; \text{Speed of light in a medium} \; (v_{B})}
  =  cvB\;=\; \frac{c}{v_{B}}
c  =nB×vB=1.5×2×108=3×108 m/s.c \;= n_{B} \times v_{B} = 1.5 \times 2 \times 10^8 = 3 \times 10^8 \text{ m/s} .
(ii) nA=  cvAn_{A} = \; \frac{c}{v_{A}}
vA=  cnA=  3×1082=1.5×108 m/sv_{A} = \; \frac{c}{n_{A}} = \; \frac{3 \times 10^8}{2}=1.5 \times 10^8 \text{ m/s}
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