A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of $μ_{g}$ would be

(\frac{4}{3}) s i n i

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\frac{1}{s i n i}

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\frac{4}{3}

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Solution

The correct option is B $\frac{1}{s i n i}$
For glass water interface $_{g} μ_{ω} = \frac{s i n i}{s i n r} . . . . . (i)$ and For water-air interface $_{ω} μ_{a} = \frac{s i n r}{s i n 90} . . . . . . (i i)$
$∴_{g} μ_{ω} x_{ω} μ_{a} = s i n i \Rightarrow μ_{g} = \frac{1}{s i n i}$

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A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of μg would be

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A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of $μ_{g}$ would be

The correct option is B $\frac{1}{s i n i}$
For glass water interface $_{g} μ_{ω} = \frac{s i n i}{s i n r} . . . . . (i)$ and For water-air interface $_{ω} μ_{a} = \frac{s i n r}{s i n 90} . . . . . . (i i)$
$∴_{g} μ_{ω} x_{ω} μ_{a} = s i n i \Rightarrow μ_{g} = \frac{1}{s i n i}$