Question

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

Answer 1

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°

Answer 2

Angle of transmitted ray is
`29.14^(o)`

Step-by-step explanation:

According to snell's law we have

`n_(1)sin(\theta _(i))=n_(2)sin(\theta r)`

Since the incident medium is air thus we have
`n_(1)=1`

By definition of refractive index we have

`n=(c)/(v)`

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

`n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2`

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

`sin(\theta _(r))=(sin(31.3)/(1.2)\\\\\therefore sin(\theta _(r))=0.4329\\\\\theta_(r)=sin^(-1)(0.4329)\\\\\theta _(r)=29.14`