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The critical angle of glass relative to ...

The critical angle of glass relative to a liquid is `57^(@)20'`. Calculate the velocity of light in the liquid. Given `mu` of glass = 1.58, Velocity of light in vacuum `3 ×10^(8) m * s^(-1) and sin 57^(@)20' = 0.8418`.

Text Solution

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Critical angle of glass relative to the liquid, `theta_( c ) = 57^(@)20.`.
If the refractive index of glass with respect to the liquid is `l^(mu)g` then,
`sintheta_(c) = (1)/(l^(mu)g) = (1)/(mu_(g)/(mu_(1))) = (mu_(1))/(mu_(g))`
`therefore " " (mu_(1))/(mu_(g)) = sin57^(@)20. = 0.8418`
`therefore " " mu_(1) = 0.8418 xx mu_(g) = 0.8418 xx 1.58`
`"again", mu_(1) = ("velocity of light in vacuum")/("velocity of light in the liquid")`
`therefore "Veolcity of the light in the liquid"`
`= (3 xx 10^(8))/(mu_(1)) = ( 3 xx 10^(8))/(0.8418 xx 1.58) = 2.255 xx 10^(8) m * s^(-1)`
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