Untitled Flashcards Set

Front (Problem Statement):
A beam of light travels through a liquid (n1=1.4n_1 = 1.4n1​=1.4) and hits a surface at θ1=52∘\theta_1 = 52^\circθ1​=52∘. It refracts into another medium at θ2=69.5∘\theta_2 = 69.5^\circθ2​=69.5∘.
Find the index of refraction of the second medium (n2n_2n2​).

Back (Solution Steps):

  • Use Snell’s Law: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2n1​sinθ1​=n2​sinθ2​.

  • Solve for n2n_2n2​: n2=n1sin⁡θ1sin⁡θ2n_2 = \frac{n_1 \sin \theta_1}{\sin \theta_2}n2​=sinθ2​n1​sinθ1​​

  • Plug in values: n2=1.4sin⁡(52∘)sin⁡(69.5∘)n_2 = \frac{1.4 \sin(52^\circ)}{\sin(69.5^\circ)}n2​=sin(69.5∘)1.4sin(52∘)​.

  • Calculate to get n2≈1.18n_2 \approx 1.18n2​≈1.18.

Flashcard 2

Front (Problem Statement):
Find the velocity of light in medium 1 (n1=1.4n_1 = 1.4n1​=1.4).

Back (Solution Steps):

  • Use the speed formula: v=cnv = \frac{c}{n}v=nc​.

  • Given c=3.00×108c = 3.00 \times 10^8c=3.00×108 m/s.

  • Solve for v1v_1v1​: v1=3.00×1081.4v_1 = \frac{3.00 \times 10^8}{1.4}v1​=1.43.00×108​

  • Calculate to get v1≈2.14×108v_1 \approx 2.14 \times 10^8v1​≈2.14×108 m/s.

Flashcard 3

Front (Problem Statement):
Find the velocity of light in medium 2 (n2=1.18n_2 = 1.18n2​=1.18).

Back (Solution Steps):

  • Use v=cnv = \frac{c}{n}v=nc​.

  • Solve for v2v_2v2​: v2=3.00×1081.18v_2 = \frac{3.00 \times 10^8}{1.18}v2​=1.183.00×108​

  • Calculate to get v2≈2.55×108v_2 \approx 2.55 \times 10^8v2​≈2.55×108 m/s.

Flashcard 4

Front (Problem Statement):
Find the speed of light in water (n=1.333n = 1.333n=1.333).

Back (Solution Steps):

  • Use v=cnv = \frac{c}{n}v=nc​.

  • Solve for vwv_wvw​: vw=3.00×1081.333v_w = \frac{3.00 \times 10^8}{1.333}vw​=1.3333.00×108​

  • Calculate to get vw≈2.25×108v_w \approx 2.25 \times 10^8vw​≈2.25×108 m/s.