CBSE Class 12 · 2019 · Wave OpticsShort Answer

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Define the term "refractive index" of a medium. Verify Snell's law of refraction when a plane wavefront is propagating from a denser to a rarer medium.

Answer: Refractive index $μ = \frac{c}{v}$ ; Huygens' construction gives $\frac{sin i}{sin r} = \frac{μ _{2}}{μ _{1}}$ .

Solution

Refractive index:

the factor by which the speed of light is reduced in a medium relative to vacuum,

μ = \frac{c}{v}

where

c

is the speed of light in vacuum and

v

in the medium.

Snell’s law (denser to rarer):

let the incident wavefront

B_{1} B_{2}

in medium 1 (speed

V_{1}

) meet the interface, and the refracted wavefront

A_{1} A_{2}

in medium 2 (speed

V_{2}

) make angle

r

. In time

T

B_{1} A_{1} = V_{1} T

and the secondary wavelet spreads

A_{2} B_{2} = V_{2} T

sin i = \frac{B _{1} A _{1}}{A _{1} B _{2}} = \frac{V _{1} T}{A _{1} B _{2}}, sin r = \frac{A _{2} B _{2}}{A _{1} B _{2}} = \frac{V _{2} T}{A _{1} B _{2}}

∴ \frac{sin i}{sin r} = \frac{V _{1}}{V _{2}} = \frac{μ _{2}}{μ _{1}}

\frac{sin i}{sin r} = \frac{μ _{2}}{μ _{1}}

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CBSE Class 12 · 2019 · Wave OpticsShort Answer

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