**Snell's Law and Refraction Example**A beam of light in water (refractive index of 1.33) enters a glass slab (refractive index of 1.50) at an angle of incidence of 60.0°. What is the angle of refraction in the glass?Options:a) 27.5° b) 39.8° c) 60.0° d) 90.0° e) 50.2°To solve this problem, use Snell's Law, which states:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]Where:- \( n_1 \) = refractive index of the first medium (water, 1.33)- \( \theta_1 \) = angle of incidence (60.0°)- \( n_2 \) = refractive index of the second medium (glass, 1.50)- \( \theta_2 \) = angle of refraction in the glassSolve for \(\theta_2\) using the formula:\[ \sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1) \]Substitute the known values to find the angle of refraction.

Write the given data Refractive index n1=1.33Refractive index n2=1.50Angle of incidence sin i=60° Solution- by using snell lawn1sin i=n2sin rn1=1.33n2=1.50sin i=60°n1sin i=n2sin r1.33sin(60°)=1.50 sin rsin r=1.33sin(60°)1.50sin r=1.33×0.8661.50sin r=0.76r=sin-1(0.76)r=50.2°Option(e) 50.2° is correct answer

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A beam of light in water (of refractive index of 1.33) enters a glass slab (c refractive index 1.50) at an angle of incidence of 60.0°. What is the angle refraction in the glass? a) 27.5° b) 39.8° c) 60.0° d) 90.0° e) 50.2°

A beam of light in water (of refractive index of 1.33) enters a glass slab (c refractive index 1.50) at an angle of incidence of 60.0°. What is the angle refraction in the glass? a) 27.5° b) 39.8° c) 60.0° d) 90.0° e) 50.2°

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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