Question

A thin flashlight beam traveling in air strikes a glass plate at an angle of 38 degrees with the closet on the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the refracted beam make with the normal in the glass?

Answer 1

The given problem can be exemplified in the following diagram:

We can use Snell's law which says the following:

`n_{\text{air}}\sin i=n_{}\sin r`

Where:

`\begin{gathered} i=\text{ angle of incidence} \\ r=\text{ angle of refraction} \\ n=\text{ index of refraction of the material} \\ n_{\text{air}}=\text{ index of refraction of air} \end{gathered}`

We will take the index of refraction of air to be 1. Now we solve for the angle of refraction:

`\sin i=n\sin r`

Now we divide by "n"

`(\sin i)/(n)=\sin r`

Taking the inverse sine function:

`\arcsin ((\sin i)/(n))=r`

The angle of incidence can be determined having into account that the sum of the given angle and the angle of incidence must be equal to 90, therefore:

`\begin{gathered} 38+i=90 \\ i=90-38 \\ i=52 \end{gathered}`

Now we substitute the values:

`\arcsin ((\sin 52)/(1.4))=r`

Solving we get:

`r=34.25`

Therefore, the angle of refraction is 34.25 degrees.