A ray of light is incident at a glass-water interface at an angle $(i_{1})$ , it emerges finally parallel to the surface of water, then the value of the refractive index of glass $(μ_{g})$ is ${μ_{w} = refractive index of water = 4/3, sin (90)^{\circ} = 1, cos (90)^{\circ} = 0}$

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Applying Snell's law at the glass-water interface:

$μ_{g} sin (i_{1}) = μ_{w} sin (r)$

Applying Snell's law at the water-air interface:

$μ_{w} sin (i_{2}) = μ_{a} sin (9 0^{\circ})$

Since the interfaces are parallel, the alternate interior angles are equal, giving $r = i_{2}$ .

Equating the expressions:

$μ_{g} sin (i_{1}) = μ_{w} sin (i_{2}) = μ_{a} sin (9 0^{\circ})$

Substituting $μ_{a} = 1$ and $sin (9 0^{\circ}) = 1$ :

$μ_{g} sin (i_{1}) = 1 \times 1 = 1$

$μ_{g} = \frac{1}{sin ( i _{1} )}$

Taking $i_{1}$ as $i$ to match the given options, we get $μ_{g} = \frac{1}{sin ( i )}$ .

Answer: $\frac{1}{sin ( i )}$

Physics - Geometrical Optics Question with Solution | TestHub

Options:

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