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a. A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 26.7° and the angle of refraction is 18.3°, Find the critical angle for the liquid when surrounded by air?b. A light ray, traveling in air, strikes the surface of abeaker of mineral oil at an angle of 37.5° with thenormal to the surface. The speed of light in mineral oil is 2.17 x10^8 m/s.. Calculate the angle of refraction.

Respuesta :

Answer:

(a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Explanation:

Given that,

Incidence angle = 26.7°

Refraction angle = 18.3°

(a). We need to calculate the refraction of liquid

Using Snell's law

[tex]n=\dfrac{\sin i}{\sin r}[/tex]

Put the value into the formula

[tex]n=\dfrac{\sin 26.7}{\sin 18.3}[/tex]

[tex]n=1.43[/tex]

We need to critical angle for the liquid when surrounded by air

Using formula of critical angle

[tex]C=\sin^{-1}(\dfrac{1}{n})[/tex]

Put the value into the formula

[tex]C=\sin^{-1}(\dfrac{1}{1.43})[/tex]

[tex]C=44.37^{\circ}[/tex]

(b). Given that,

Incidence angle = 37.5°

Speed of light in mineral [tex]v=2.17\times10^{8}\ m/s[/tex]

We need to calculate the index of refraction

Using formula of index of refraction

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

Put the value into the formula

[tex]n=\dfrac{3\times10^{8}}{2.17\times10^{8}}[/tex]

[tex]n=1.38[/tex]

We need to calculate the angle of refraction

Using Snell's law

[tex]n=\dfrac{\sin i}{\sin r}[/tex]

[tex]\sin r=\dfrac{\sin i}{n}[/tex]

Put the value into the formula

[tex]\sin r=\dfrac{\sin 37.5}{1.38}[/tex]

[tex]r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})[/tex]

[tex]r=26.17^{\circ}[/tex]

Hence, (a). The critical angle for the liquid when surrounded by air is 44.37°

(b). The angle of refraction is 26.17°.

Answer

a) Angle of incidence i  = 26.7°

Angle of refraction r = 18.3°

From Snell’s law index of refraction of the liquid

[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]

[tex]n = \dfrac{sin\ 26.7^0}{sin\ 18.3^0}[/tex]

      n = 1.43

So, critical angle

[tex]C= sin^{-1}(\dfrac{1}{n})[/tex]

[tex]C= sin^{-1}(\dfrac{1}{1.43})[/tex]

       C = 44.33°

b) Angle of incidence, i = 37.5°

  speed of light in mineral oil , v = 2.17 x 10⁸ m/s

  speed of light in air, c = 3 x 10⁸ m/s

refractive index of the oil

 [tex]n = \dfrac{c}{v}[/tex]

 [tex]n = \dfrac{3\times 10^8}{2.17\times 10^8}[/tex]

  n = 1.38

again using Snell's law

[tex]n = \dfrac{sin\ i}{sin\ r}[/tex]

[tex]sin\ r = \dfrac{sin\ i}{n}[/tex]

[tex]sin\ r = \dfrac{sin\ 37.5^0}{1.38}[/tex]

[tex] r = sin^{-1}(0.441)[/tex]

       r = 26.18°

hence, the angle of refraction is equal to r = 26.18°