The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

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A1peakenbe A1peakenbe · Answer 1 · 2019-09-06T11:56:31+02:00

Answer:

Angle of transmitted ray is [tex]29.14^{o}[/tex]

Explanation:

According to snell's law we have

[tex]n_{1}sin(\theta _{i})=n_{2}sin(\theta r)[/tex]

Since the incident medium is air thus we have [tex]n_{1}=1[/tex]

By definition of refractive index we have

[tex]n=\frac{c}{v}[/tex]

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

[tex]n_{2}=\frac{3\times 10^{8}}{2.5\times 10^{8}}=1.2[/tex]

Thus using the calculated values in Snell's law we obtain

[tex]sin(\theta _{r})=\frac{sin(31.3}{1.2}\\\\\therefore sin(\theta _{r})=0.4329\\\\\theta_{r}=sin^{-1}(0.4329)\\\\\theta _{r}=29.14[/tex]

BatteringRam BatteringRam · Answer 2 · 2019-09-06T17:35:47+02:00

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i

n₁ sin i = n₂ sin r

⇒ sin r = (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°