A ray of light is incident at the glass-water interface at an angle $i$ . It merges finally parallel to the surface of the water. Then, the value of $μ_{g}$ would be(where $μ_{g}$ is the refractive index of glass with respect to water)

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Answer 1

Applying Snell's law at glass-water surface,
$_{g}^{w} μ = \frac{sin r}{sin i} = \frac{_{g} μ}{_{w} μ} . . . (i)$
Applying Snell's law at water-air surface,
$_{w}^{a} μ = \frac{sin 9 0 ^{\circ}}{sin r} = \frac{_{w} μ}{_{a} μ} . . . (i i)$
Multiplying (i) and (ii),
$\frac{_{g} μ}{_{a} μ} = \frac{1}{sin i} \Rightarrow_{g} μ = \frac{1}{sin i}$ .

Answer 2

Applying Snell's law at glass-water surface,

_{g}^{w} μ = \frac{sin r}{sin i} = \frac{_{g} μ}{_{w} μ} . . . (i)

Applying Snell's law at water-air surface,

_{w}^{a} μ = \frac{sin 9 0 ^{\circ}}{sin r} = \frac{_{w} μ}{_{a} μ} . . . (i i)

Multiplying (i) and (ii),

\frac{_{g} μ}{_{a} μ} = \frac{1}{sin i} \Rightarrow_{g} μ = \frac{1}{sin i}

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A ray of light is incident at the glass-water interface at an angle $i$ . It merges finally parallel to the surface of the water. Then, the value of $μ_{g}$ would be(where $μ_{g}$ is the refractive index of glass with respect to water)

$(4 / 3) sin i$

$1 / sin i$

$4 / 3$

$1$

Critical angle of glass is $θ_{1}$ and that of water is $θ_{2}$ . The critical angle for water and glass surface would be $(μ_{g} = 3 / 2, μ_{w} = 4 / 3)$ .

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 :

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

A ray of light $(G H)$ is incident on the glass-water interface $D C$ at an angle $i$ . It emerges in air along the water air interface $E F$ (see figure). If the refractive index of water $μ_{w}$ is $4 / 3$ , the refractive index of glass $μ_{g}$ is

The angle between reflected and refracted beams is $9 0^{0}$ in the water air interface. The angle of incidence in water is ( $μ_{w a t e r} = \frac{4}{3}, μ_{a i r} = 1$ )

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

A ray of light is incident at the glass-water interface at an angle i. It merges finally parallel to the surface of the water. Then, the value of μg​ would be(where μg​ is the refractive index of glass with respect to water)

(4/3)sini

1/sini

4/3

1

Applying Snell's law at glass-water surface, gw​μ=sinisinr​=w​μg​μ​...(i) Applying Snell's law at water-air surface, wa​μ=sinrsin90∘​=a​μw​μ​...(ii) Multiplying (i) and (ii), a​μg​μ​=sini1​ ⇒g​μ=sini1​.

Critical angle of glass is θ1​ and that of water is θ2​. The critical angle for water and glass surface would be (μg​=3/2,μw​=4/3).

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 :

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

A ray of light (GH) is incident on the glass-water interface DC at an angle i. It emerges in air along the water air interface EF (see figure). If the refractive index of water μw​ is 4/3, the refractive index of glass μg​ is

The angle between reflected and refracted beams is 900 in the water air interface. The angle of incidence in water is (μwater​=34​,μair​=1)

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

A ray of light is incident at the glass-water interface at an angle $i$ . It merges finally parallel to the surface of the water. Then, the value of $μ_{g}$ would be(where $μ_{g}$ is the refractive index of glass with respect to water)

$(4 / 3) sin i$

$1 / sin i$

$4 / 3$

$1$

Critical angle of glass is $θ_{1}$ and that of water is $θ_{2}$ . The critical angle for water and glass surface would be $(μ_{g} = 3 / 2, μ_{w} = 4 / 3)$ .

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 :

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

A ray of light $(G H)$ is incident on the glass-water interface $D C$ at an angle $i$ . It emerges in air along the water air interface $E F$ (see figure). If the refractive index of water $μ_{w}$ is $4 / 3$ , the refractive index of glass $μ_{g}$ is

The angle between reflected and refracted beams is $9 0^{0}$ in the water air interface. The angle of incidence in water is ( $μ_{w a t e r} = \frac{4}{3}, μ_{a i r} = 1$ )