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

\frac{4}{3} sin i

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\frac{1}{sin i}

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\frac{4}{3}

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1

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Solution

The correct option is B $\frac{1}{sin i}$
$s i n r \times μ_{ω} = μ_{a} \times 1$

$s i n r = \frac{3}{4}$

$s i n i \times μ_{G} = 1$

$μ_{G} = \frac{1}{s i n i}$

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

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

The correct option is B 1sinisin r×μω=μa×1sinr=34sini×μG=1μG=1sin i

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

The correct option is B $\frac{1}{sin i}$
$s i n r \times μ_{ω} = μ_{a} \times 1$

$s i n r = \frac{3}{4}$

$s i n i \times μ_{G} = 1$

$μ_{G} = \frac{1}{s i n i}$