Question

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.

Answer 1

Answer

a) Angle of incidence i  = 26.7°

Angle of refraction r = 18.3°

From Snell’s law index of refraction of the liquid

$n = \dfrac{sin\ i}{sin\ r}$

$n = \dfrac{sin\ 26.7^0}{sin\ 18.3^0}$

      n = 1.43

So, critical angle

$C= sin^{-1}(\dfrac{1}{n})$

$C= sin^{-1}(\dfrac{1}{1.43})$

       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

 $n = \dfrac{c}{v}$

 $n = \dfrac{3\times 10^8}{2.17\times 10^8}$

  n = 1.38

again using Snell's law

$n = \dfrac{sin\ i}{sin\ r}$

$sin\ r = \dfrac{sin\ i}{n}$

$sin\ r = \dfrac{sin\ 37.5^0}{1.38}$

$r = sin^{-1}(0.441)$

       r = 26.18°

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

Answer 2

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

$n=\dfrac{\sin i}{\sin r}$

Put the value into the formula

$n=\dfrac{\sin 26.7}{\sin 18.3}$

$n=1.43$

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

Using formula of critical angle

$C=\sin^{-1}(\dfrac{1}{n})$

Put the value into the formula

$C=\sin^{-1}(\dfrac{1}{1.43})$

$C=44.37^{\circ}$

(b). Given that,

Incidence angle = 37.5°

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

We need to calculate the index of refraction

Using formula of index of refraction

$n=\dfrac{c}{v}$

Put the value into the formula

$n=\dfrac{3\times10^{8}}{2.17\times10^{8}}$

$n=1.38$

We need to calculate the angle of refraction

Using Snell's law

$n=\dfrac{\sin i}{\sin r}$

$\sin r=\dfrac{\sin i}{n}$

Put the value into the formula

$\sin r=\dfrac{\sin 37.5}{1.38}$

$r=\sin^{-1}(\dfrac{\sin 37.5}{1.38})$

$r=26.17^{\circ}$

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

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