题目
题目

Physics 1 Hon - Bowles - p06/p07 P10.2 - Lesson 4 Exit Ticket

简答题

A ray of light in air is approaching the boundary with water (n = 1.333)  at an angle of 37.1 degrees. Determine the angle of refraction of the light ray. 

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标准答案
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思路分析
We are given light transitioning from air into water, so apply Snell's law: n1 sin(theta1) = n2 sin(theta2). Here n1 is the refractive index of air, approximately 1.000, and n2......Login to view full explanation

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Light exits a diamond (n = 2.42) into air. The angle of incidence is 15.0°. What is the angle of refraction?

Question text1. Assume that we have a planet with a spherically uniform P wave speed distribution. The P wave speed at any depth is given by: [math: v(r)=8+16(1−(r/R)2)] Where [math: r] is the distance from the core of the planet, and [math: R=8200km] is the radius of the planet. a) A P wave is generated at the surface at an angle [math: θ0=32∘] from the surface normal. Calculate to three significant figures, in degrees, the angle of refraction at a depth of 110 km. Hint: Snell's law in a spherically symmetric Earth is given by [math: r1sin(θ1)v1=r2sin(θ2)v2] Refraction angle = Answer 1 Question 8[input] degrees [4] b) Calculate the maximum depth at which the ray travels, in km to three significant figures. Hint: At the maximum depth you will have a refraction angle of [math: 90∘].Maximum depth = Answer 2 Question 8[input] km [6]

Question at position 1 Given the figure below where light travels from one medium to another, and with Snell's Law in mind as well as the difference between the angles apparent; what is the relationship between the refractive indices n_1 and n_2? n_1 > n_2n_1 < n_2n_1 = n_2Not enough information

A light ray passes from water (refractive index ( n1 = 1.33 )) into diamond (refractive index ( n2 = 2.42 )). If the angle of incidence in water is ( 300 ), what is the angle of refraction in diamond?

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