A ray of light is incident the glass-water interface an angle i- emerges finally parallel to the surface of water- then the value of refractive index of glass- mu -g- would be -dfrac-4-3- sin idfrac-1- sin i-dfrac-4-3-1

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Toppr Admin · Accepted Answer

Sin-xA0-r-xD7-x3BC-x3C9-x3BC-a-xD7-1sinr-34sini-xD7-x3BC-G-1-x3BC-G-1sin-xA0-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 :

$\frac{4}{3} sin i$
$\frac{1}{sin i}$
$\frac{4}{3}$
$1$

$s i n r \times μ_{ω} = μ_{a} \times 1$

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

$s i n i \times μ_{G} = 1$

$μ_{G} = \frac{1}{s i n i}$

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 :43sini1sini431

sin 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 :

$\frac{4}{3} sin i$
$\frac{1}{sin i}$
$\frac{4}{3}$
$1$

$s i n r \times μ_{ω} = μ_{a} \times 1$

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

$s i n i \times μ_{G} = 1$

$μ_{G} = \frac{1}{s i n i}$