Mary Hammonds

2022-01-10

A parallel beam of light in air makes an angle of ${47.5}^{\circ }$ with the surface of a glass plate having a refractive index of 1.66
a) What is the angle between the reflected part of the beam and the surface of the glass?
b) What is the angle between the refracted beam and the surface of the glass?

Esther Phillips

Step 1
Law of reflection:
${\theta }_{r}={\theta }_{a}$
${\theta }_{r}⇒\text{The angle of reflection which is measured from normal}$
${\theta }_{a}⇒\text{The angle of incidence which is measured from normal}$
Note: The normal is a line drawn perpendicular to the surface at the point where the incident ray strikes the surface.
When a light wave strikes a smooth interface separating two transparent materials (such as air and glass), the wave is generally partly refracted and partly reflected. Experiments has shown that, for smooth surfaces in reflection, the angle of incidence equals the angle of reflection as stated in the above law.
Step 2
a) The same angle of incidence equals the angle of reflection. Therefore, the incident beam makes the same angle with glass as the reflected one
$\theta ={47.5}^{\circ }⇒\text{with the glass}$
Step 3
Index of refraction:
$n=\frac{c}{v}$
$n⇒\text{index of refraction of an optical material.}$

Light always travels more slowly in a material than in vacuum, so the value of n in anything other than vacuum is always greater than unity.
Note that: the index of refraction depends on each material for the same wavelength of a wave.
Apply: in most cases we need the speed of light in a specific medium, so we use the index of refraction for this medium to get the speed of light in this medium.
Apply: In most problems, we use the index of refraction for a certain material to get the speed of light in this material. In another problems, we do the opposite to get the index of refraction for a certain material by knowing the speed of light in this material.
Step 4
b) ${\theta }_{1}={90}^{\circ }-{47.5}^{\circ }={42.5}^{\circ }$
$⇒\text{note: we want the angle with the normal and not with the surface}$
${n}_{1}{\mathrm{sin}\theta }_{1}={n}_{2}{\mathrm{sin}\theta }_{2}$
$1×{\mathrm{sin}42.5}^{\circ }=1.66{\mathrm{sin}\theta }_{2}$
${\theta }_{2}=\mathrm{arcsin}\left(\frac{1×{\mathrm{sin}42.5}^{\circ }}{1.66}\right)={24.0154}^{\circ }⇒\text{with sormal}$
$\theta ={90}^{\circ }-{24.0154}^{\circ }={66}^{\circ }⇒\text{with the surface of glass}$

abonirali59

Step 1
The angle of incidence (measured with regard to the perpendicular to the interface between the two mediums) and the angle of reflection (again, measured with respect to the interface between the two mediums), according to the law of reflection, are equal:
${\theta }_{i}={\theta }_{r}$
In our problem, ${47.5}^{\circ }$ is the angle formed by the incident ray and the surface; as a result, the angle of incidence (relative to the perpendicular) is ${\theta }_{i}={90}^{\circ }-{47.5}^{\circ }={42.5}^{\circ }$
Consequently, the reflection's angle is also ${\theta }_{r}={42.5}^{\circ }$, and because this is calculated in relation to the surface's perpendicular, the angle between the reflected ray and the surface is
$\theta ={90}^{\circ }-{42.5}^{\circ }={47.5}^{\circ }$
Step 2
b) The angle of refraction, ${\theta }_{t}$ is given by Snells

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