Question
In the adjoining diagram, a wavefront AB, moving in air is incident on a plane glass surface XY. Its position CD after refraction through a glass slab is shown also along with the normals drawn at A and D. The refractive index of glass with respect to air (μ=1 ) will be equal to 
Answer: Option B
:
B
in the case of refraction is CD is the refracted waves front and V1 and V2 are the speed of light in
the two media, then in the time the wavelets from B reaches C, the wavelets from A will reach D,
such that
t=BCva=ADvg⇒BCAD=vavg...(i)
But in △ACB,BC=ACsinθ...(ii)
While in △ACD,AD=ACsinϕ′...(iii)
From equations (i), (ii) and (iii) vavg=sinθsinϕ′
Also μ∝1v⇒vavg=μgμa=sinθsinϕ′⇒μg=sinθsinϕ′
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:
B
in the case of refraction is CD is the refracted waves front and V1 and V2 are the speed of light in
the two media, then in the time the wavelets from B reaches C, the wavelets from A will reach D,
such that
t=BCva=ADvg⇒BCAD=vavg...(i)
But in △ACB,BC=ACsinθ...(ii)
While in △ACD,AD=ACsinϕ′...(iii)
From equations (i), (ii) and (iii) vavg=sinθsinϕ′
Also μ∝1v⇒vavg=μgμa=sinθsinϕ′⇒μg=sinθsinϕ′
Was this answer helpful ?
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