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Explain the relationship between optical density and the refractive index of a medium. What does a higher refractive index indicate?

Physics

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Answer

b

25.37∘\text{25.37}^\circ

Step-by-step solution

4 steps

Part a

Snell's law states the relationship between the angles of incidence and refraction and the refractive indices of the two media.

n1sin⁡(θ1)=n2sin⁡(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2)

where n1n_1 is the refractive index of the first medium, θ1\theta_1 is the angle of incidence, n2n_2 is the refractive index of the second medium, and θ2\theta_2 is the angle of refraction.

Part b

25.37∘\text{25.37}^\circ

  1. State the given values.

    Given:
    Refractive index of air, n1=1.0n_1 = 1.0
    Angle of incidence in air, θ1=40∘\theta_1 = 40^\circ
    Refractive index of glass, n2=1.5n_2 = 1.5

  2. Apply Snell's Law.

    n1sin⁡(θ1)=n2sin⁡(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2) 1.0×sin⁡(40∘)=1.5×sin⁡(θ2)1.0 \times \sin(40^\circ) = 1.5 \times \sin(\theta_2)

  3. Calculate sin⁡(θ2)\sin(\theta_2).

    0.6428=1.5×sin⁡(θ2)0.6428 = 1.5 \times \sin(\theta_2) sin⁡(θ2)=0.64281.5\sin(\theta_2) = \frac{0.6428}{1.5} sin⁡(θ2)≈0.4285\sin(\theta_2) \approx 0.4285

  4. Calculate θ2\theta_2.

    θ2=arcsin⁡(0.4285)\theta_2 = \arcsin(0.4285) θ2≈25.37∘\theta_2 \approx 25.37^\circ

    The angle of refraction in the glass is 25.37∘\boxed{\text{25.37}^\circ}.

Part c

The light ray will bend towards the normal. This is because light is traveling from a medium with a lower refractive index (air, n=1.0n=1.0) to a medium with a higher refractive index (glass, n=1.5n=1.5). When light enters a denser optical medium, it slows down and bends towards the normal.

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