A ray of light travelling in glass - - 3-2 is incident on a horizontal glass-air surface at the critical angle -C- If a thin layer of water - - 4-3 is now poured on the glass-air surface- the angle at which the ray emerges into air at the water-air surface is

The correct option is B 90^∘ μ_gsinθ_c = μ_1sin 90^∘ or μ_gsinθ_c = 1 When water is poured, μ_wsin r = μ_gsinθ_c or μ_wsin r = 1 Again, ...

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Standard X

Physics

Critical Angle

A ray of ligh...

Question

A ray of light travelling in glass $(μ = 3 / 2)$ is incident on a horizontal glass-air surface at the critical angle $θ_{C}$ . If a thin layer of water $(μ = 4 / 3)$ is now poured on the glass-air surface, the angle at which the ray emerges into air at the water-air surface is

60^{\circ}

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45^{\circ}

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90^{\circ}

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180^{\circ}

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Solution

The correct option is B $90^{\circ}$
$μ_{g} sin θ_{c} = μ_{1} sin 90^{\circ}$
or $μ_{g} sin θ_{c} = 1$
When water is poured,
$μ_{w} sin r = μ_{g} sin θ_{c}$
or $μ_{w} sin r = 1$
Again, $μ_{a} sin θ = μ_{w} sin r$
or $μ_{a} sin θ = 1$
or $sin θ = 1$ or $θ = 90^{\circ}$ .

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A ray of light travelling in glass (μ=3/2) is incident on a horizontal glass-air surface at the critical angle θC. If a thin layer of water (μ=4/3) is now poured on the glass-air surface, the angle at which the ray emerges into air at the water-air surface is

The correct option is B 90∘μgsinθc=μ1sin90∘or μgsinθc=1When water is poured,μwsinr=μgsinθcor μwsinr=1Again, μasinθ=μwsinror μasinθ=1or sinθ=1 or θ=90∘.

A ray of light travelling in glass $(μ = 3 / 2)$ is incident on a horizontal glass-air surface at the critical angle $θ_{C}$ . If a thin layer of water $(μ = 4 / 3)$ is now poured on the glass-air surface, the angle at which the ray emerges into air at the water-air surface is

The correct option is B $90^{\circ}$
$μ_{g} sin θ_{c} = μ_{1} sin 90^{\circ}$
or $μ_{g} sin θ_{c} = 1$
When water is poured,
$μ_{w} sin r = μ_{g} sin θ_{c}$
or $μ_{w} sin r = 1$
Again, $μ_{a} sin θ = μ_{w} sin r$
or $μ_{a} sin θ = 1$
or $sin θ = 1$ or $θ = 90^{\circ}$ .