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The figure below shows a ray of light passing from glass to air at an angle of 60° to the interface. The refractive index of glass is 1.5, determine the angle of refraction.

Physics

ScanSolve AI Answer

Answer

48.59∘\text{48.59}^\circ

Step-by-step solution

5 steps
  1. Tambua thamani ulizopewa na kanuni.

    Tunaambiwa kuwa mwale wa mwanga unapita kutoka kioo kwenda hewani.
    Kielezo cha mtawanyiko wa kioo, n1=1.5n_1 = 1.5.
    Kielezo cha mtawanyiko wa hewa, n2=1.0n_2 = 1.0.
    Pembe ya tukio kwa kiolesura ni 60∘60^\circ. Hii inamaanisha pembe ya tukio (angle of incidence) na mstari wa kawaida (normal) ni 90∘−60∘=30∘90^\circ - 60^\circ = 30^\circ. Hivyo, θ1=30∘\theta_1 = 30^\circ.
    Tutatumia Kanuni ya Snell: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2.

  2. Badilisha thamani kwenye Kanuni ya Snell.

    1.5×sin⁡(30∘)=1.0×sin⁡θ21.5 \times \sin(30^\circ) = 1.0 \times \sin \theta_2
  3. Kokotoa sin⁡(30∘)\sin(30^\circ).

    Tunajua kuwa sin⁡(30∘)=0.5\sin(30^\circ) = 0.5.

    1.5×0.5=1.0×sin⁡θ21.5 \times 0.5 = 1.0 \times \sin \theta_2
  4. Rahisisha na utafute sin⁡θ2\sin \theta_2.

    0.75=sin⁡θ20.75 = \sin \theta_2
  5. Tafuta pembe ya mtawanyiko, θ2\theta_2.

    Ili kupata θ2\theta_2, tunatumia kazi ya arcsin (au sin⁡−1\sin^{-1}).

    θ2=arcsin⁡(0.75)\theta_2 = \arcsin(0.75) θ2≈48.59∘\theta_2 \approx 48.59^\circ

    Pembe ya mtawanyiko ni 48.59∘\text{48.59}^\circ.

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