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A ray of light of vacuum wavelength 550 nm traveling in air enters a slab of transparent material. The incoming ray makes an angle of 40.0° with the normal, and the refracted ray makes an angle of 26.0° with the normal. Find the index of refraction of the transparent material. (Assume that the index of refraction of air for light of wavelength 550 nm is 1.00.)​

Sagot :

Answer:

sin 40 = n sin 26    snell's law for incoming ray in air

n = sin 40 / sin 26 = 1.47

The index of refraction of the transparent material is 1.47. The value of the refractive index depends upon the material.

What is an index of refraction?

The refractive index of a substance also known as the refraction index or index of refraction is a dimensionless quantity that specifies how quickly light passes through an object.

The given data in the problem is;

The wavelength of a ray of light in a vacuum is,  [tex]\lambda = 550 \ nm[/tex]

The incoming ray makes an angle,   [tex]\rm \theta = 40^0[/tex]

the refracted ray makes an angle,[tex]\rm \theta_R= 26.0^0[/tex]

The index of refraction of the transparent material is,n

The index of refraction is found as;

[tex]\rm n= \frac{sin 40^0}{sin 26^0} \\\\ n= 1.47[/tex]

Hence the index of refraction of this medium will be 1.47.

To learn more about the index of refraction refer to the link;

https://brainly.com/question/23750645