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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsTIR/Prism and DispersionMedium2 minQB

PhysicsMediummultiple choice

A cubical block of glass of refractive index $n_{1}$ is in contact with the surface of water of refractive index $n_{2}$ . A ray of light is incident on vertical face of the block and after refraction, a total internal reflection at the base and refraction at the opposite vertical face, the ray emerges out at an angle $θ$ . Then :-

Options:(select one or more)

Answer:

Solution:

At entry: $n_{1} sin ϕ = sin θ$ , so $cos ϕ = \frac{n _{1}^{2} - s i n ^{2} θ}{n _{1}}$ .

TIR at the base needs $9 0^{\circ} - ϕ > C$ , i.e. $cos ϕ > sin C = \frac{n _{2}}{n _{1}}$ .

So $n_{1}^{2} - sin^{2} θ > n_{2}^{2}$ , giving $sin^{2} θ < n_{1}^{2} - n_{2}^{2}$ and this requires $n_{2} < n_{1}$ . Hence (A, C).

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:TIR/Prism and Dispersion

⏱ 2mℹ️ Source: QB

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