To solve this problem it is necessary to apply the concepts related to the Snell Law of refraction.
Snell's law tells us that
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
Where,
[tex]n_{1,2}[/tex] = Refractive index for the material
[tex]\theta_{1,2}[/tex] = Angle of refraction
For this particular case we have that
[tex]n_1 = 1.64[/tex]
[tex]n_2 = 1.50[/tex]
[tex]\theta_2 = 90\° \rightarrow[/tex] The minimum angle and the maximum angle must be used.
Substituting,
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
[tex]1.64*sin\theta_1=1.5*sin90[/tex]
[tex]\theta = sin^{-1}(\frac{1.5}{1.64})[/tex]
[tex]\theta = 66.15\°[/tex]
The maximum angle will be calculated by substracting the value of minimum angle (90°)
[tex]\theta_{max} = 90-66.15[/tex]
[tex]\theta_{max} = 23.846\°[/tex]
Therefore the maximum angle that a light ray can make with the wall of the core is 23.84°
The maximum angle between a light ray and the wall of the core is 23.85°.
Given the following data:
Refractive index of optical fiber = 1.64.
Refractive index of cladding = 1.50.
Snell's law gives the relationship between the angle of incidence and angle of refraction with respect to light or other waves that are passing through a media or two different substances such as glass, water, cladding, or air.
Snell's Law states that the when light travels from one medium to another, it generally refracts. Mathematically, it is given by this formula:
[tex]n_1sin \theta_1 = n_2sin \theta_2\\\\1.64sin \theta_1 = 1.50sin 90\\\\1.64sin \theta_1 =1.50\\\\sin \theta_1 =\frac{1.50}{1.64} \\\\\theta_1= sin^{-1}(0.9146)\\\\[/tex]
Angle 1 = 66.15°.
For the maximum angle:
[tex]\Theta _{max}=90-66.15[/tex]
Maximum angle = 23.85°.
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