Physics, asked by MathematicalNerd, 3 months ago

The refractive index of a denser medium with respect to a rarer medium is 1.6. The difference between the velocities of light in the two mediums are 0.75*10^8 m/s. Find the velocities of light in the two mediums and their refractive indices. (C=3*10^8 m/s)

Answers

Answered by sonurahulsai
1

Answer:

velocities is the speed of the motion of the velocity

Answered by varsha6033
1

Answer:

Refractive index of medium 1 with respect to medium 2 is velocity of light in medium 2 divided by velocity of light in medium 1.

The phenomenon of change of velocity of light at the interface between two media is called refraction.

n12(read as n of 1 w.r.t.2)=v2/v1=(f.Lambda2)/(f.Lambda 1)=(Lambda2)/(Lambda1)=sin i/sin r. Here f is frequency of light which remains constant due to law of conservation of energy. Here, i is angle of incidence formed at interface in medium 2 and r is angle of refraction in madium 1.

According to Snell’s law n2sin i=n1 sin r. Therefore sin i/sin r=n1/n2 or n12=ni/n2. Here, n1 and n2 are absolute refractive indices of medium 1 and 2. Absolute refractive index means refractive index w.r.t. air or vacuum.

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