/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A glass lens has a scratch-resis... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 16

A glass lens has a scratch-resistant plastic coating on it. The speed of light in the glass is \(0.67 c\) and the speed of light in the coating is \(0.80 c .\) A ray of light in the coating is incident on the plastic-glass boundary at an angle of \(12.0^{\circ}\) with respect to the normal. At what angle with respect to the normal is the ray transmitted into the glass?

Short Answer

Expert verified
Answer: The angle at which the ray of light is transmitted into the glass after passing through the coating is approximately \(10.10^{\circ}\).

Step by step solution

01

Compute the refractive indices

First, we need to find the refractive indices of the glass and the coating. The refractive index is the ratio of the speed of light in vacuum to the speed of light in the material. We are given that the speed of light in the glass is \(0.67c\) and in the coating is \(0.80c.\) Using these speeds, we can compute their refractive indices: Refractive Index of glass (n1) = \(\frac{c}{0.67c}\) = \(\frac{1}{0.67}\) Refractive Index of coating (n2) = \(\frac{c}{0.80c}\) = \(\frac{1}{0.80}\)
02

Apply Snell's Law

Now, we'll apply Snell's Law to find the angle of refraction in the glass. Snell's Law states the relationship between angles of incidence and refraction as: \(n_1\sin{\theta_1} = n_2\sin{\theta_2}\) Here, \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively. We are given that the angle of incidence \(\theta_1 = 12^{\circ}.\) So, \(\sin{\theta_1} = \sin{12^{\circ}}.\) We can now solve Snell's Law for \(\theta_2:\) \(\frac{1}{0.67}\sin{12^{\circ}} = \frac{1}{0.80}\sin{\theta_2}\) \(\sin{\theta_2} = \frac{0.67}{0.80}\sin{12^{\circ}}\)
03

Calculate the angle of transmission

Now, we'll find the angle \(\theta_2\) by taking the inverse sine of the result: \(\theta_2 = \arcsin{\left(\frac{0.67}{0.80}\sin{12^{\circ}}\right)}\) Calculating the value for this expression, we obtain: \(\theta_2 \approx 10.10^{\circ}\) So, the angle at which the ray of light is transmitted into the glass with respect to the normal is approximately \(10.10^{\circ}\).

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Most popular questions from this chapter

A diamond in air is illuminated with white light. On one particular facet, the angle of incidence is \(26.00^{\circ} .\) Inside the diamond, red light \((\lambda=660.0 \mathrm{nm}\) in vacuum) is refracted at \(10.48^{\circ}\) with respect to the normal; blue light $(\lambda=470.0 \mathrm{nm} \text { in vacuum })\( is refracted at \)10.33^{\circ} .$ (a) What are the indices of refraction for red and blue light in diamond? (b) What is the ratio of the speed of red light to the speed of blue light in diamond? (c) How would a diamond look if there were no dispersion?
The angle of deviation through a triangular prism is defined as the angle between the incident ray and the emerging ray (angle \(\delta\) ). It can be shown that when the angle of incidence \(i\) is equal to the angle of refraction \(r^{\prime}\) for the emerging ray, the angle of deviation is at a minimum. Show that the minimum deviation angle \(\left(\delta_{\min }=D\right)\) is related to the prism angle \(A\) and the index of refraction \(n,\) by $$ n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A} $$ [Hint: For an isosceles triangular prism, the minimum angle of deviation occurs when the ray inside the prism is parallel to the base, as shown in the figure.]
A \(1.80-\mathrm{cm}\) -high object is placed \(20.0 \mathrm{cm}\) in front of a concave mirror with a \(5.00-\mathrm{cm}\) focal length. What is the position of the image? Draw a ray diagram to illustrate.

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The focal length of a thin lens is \(-20.0 \mathrm{cm} .\) A screen is placed \(160 \mathrm{cm}\) from the lens. What is the \(y\) -coordinate of the point where the light ray shown hits the screen? The incident ray is parallel to the central axis and is \(1.0 \mathrm{cm}\) from that axis.
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