Physics, 13.11.2019 11:31 emily200705

Aray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with the normal. part of the light is reflected and part is refracted. find the angle between the reflected and the refracted light. answer in units of ◦ .

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Answer from: Sk8terkaylee

$135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium refractive index (the air)

$n_{2}=1.309$ is the second medium refractive index (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the angle between the reflected and the refracted light.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$ (2)

Remembering that the Normal makes a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$ (3)

Finding $\beta$ :

$\beta=90\º-25.5\º=64.5\º$ (4)

Doing the same with $\theta_{2}$ and $\alpha}$ :

$90\º=\theta_{2}+\alpha$ (5)

Finding $\alpha$ :

$\alpha=90\º-19.201\º=70.79\º$ (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$ (7)

$\alpha+\beta=135.29\º$ >>>This is the angle between the reflected and the refracted light.

Aray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) a

Answer from: Quest

Ibelieve your answer is b.hope this : )

Answer from: Quest

The answer is 4f======since electrostatic force is given by $f_{elec} = \dfrac{kq_1 q_2}{r^2}$ we let f be the electrostatic force when r = 1 m $f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2$ if we let $f_{0.5}$ denote the electrostatic force when the distance is decreased so that r = 0.5m, then $\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}$ we established earlier that f = kq₁q₂. therefore $f_{0.5} = 4 f$ the force will now be 4f.

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