Question

As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated.
a) Using the information on the figure, find the index of refraction of material X .
b) Find the angle the light makes with the normal in the air.

Answer 1

The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.

In order to determine the solution, we must understand Snell's law.

What is the refraction law of Snell? How can the issue be resolved with this?

• One way to express Snell's law for refraction is as follows:

                            $\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}$

where the refractive index is n and the incidence angle is i. The refracted angle is r.

• As is well known, water has a refractive index of 1.33.

• The incidence angle from the image in the first case is 65°, while the refracted angle is 48°. Consequently, the medium X's refractive index will be,

                      $n_X=\frac{n_w*sin 48}{sin65} =1.09$

• The incident angle in the second scenario is 48°, and we must determine the refracted angle r for the air.

• Since we now know that air has a refractive index of 1, so that the refracted angle is,

                  $sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.$

As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.

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Answer 2

Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Explanation: To find the answer, we need to know the Snell's law.

What is Snell's law of refraction? Using this, how to solve the problem?

• The Snell's law for refraction can be written as,

                      $\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}$

where, i is the incident angle, r is the refracted angle, n is the refractive index.

• As we know that the refractive index of water is 1.33
• For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,

                           $\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09$

• In the second case, incident angle is 48° and we have to find the refracted angle r for the air.
• As we know that the refractive index of air is 1.
• Thus, the refracted angle will be,

                         $\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.$

Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Learn more about the Snell's law here:

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