Math Problem Statement

2024-25 S3 Physics Wave Motion I Homework 2: Questions on light refraction involving Snell's Law and refractive indices, covering the transition of light between air, glass, and liquids.

Solution

I'll go over each problem and verify or solve based on what's visible.

Problem Analysis and Solutions

  1. Problem 1: A ray of light passes from a liquid to air. The angles of incidence and refraction are 35° and 53° respectively. We need to find the refractive index of the liquid.

    Using Snell's Law: nliquidsin(35)=nairsin(53)n_{\text{liquid}} \sin(35^\circ) = n_{\text{air}} \sin(53^\circ) Assuming nair=1n_{\text{air}} = 1, nliquid=sin(53)sin(35)n_{\text{liquid}} = \frac{\sin(53^\circ)}{\sin(35^\circ)} Calculating this: nliquid0.79860.57361.39n_{\text{liquid}} \approx \frac{0.7986}{0.5736} \approx 1.39 Answer: 1.39

  2. Problem 2: A light ray travels from air to glass with angles of incidence and refraction given as 25° and 50° respectively. We need to find the refractive index of the glass.

    Using Snell's Law: nairsin(25)=nglasssin(50)n_{\text{air}} \sin(25^\circ) = n_{\text{glass}} \sin(50^\circ) Assuming nair=1n_{\text{air}} = 1, nglass=sin(25)sin(50)n_{\text{glass}} = \frac{\sin(25^\circ)}{\sin(50^\circ)} Calculating this: nglass0.42260.76601.44n_{\text{glass}} \approx \frac{0.4226}{0.7660} \approx 1.44 Answer: 1.44

  3. Problem 3: A light ray travels from glass to air, with an angle of refraction of 55°. The refractive index of glass is given as 1.5. We need to find the angle of incidence.

    Using Snell's Law: nglasssin(θincident)=nairsin(55)n_{\text{glass}} \sin(\theta_{\text{incident}}) = n_{\text{air}} \sin(55^\circ) Solving for θincident\theta_{\text{incident}}: sin(θincident)=sin(55)1.5\sin(\theta_{\text{incident}}) = \frac{\sin(55^\circ)}{1.5} Calculating this: sin(θincident)0.81921.50.5461\sin(\theta_{\text{incident}}) \approx \frac{0.8192}{1.5} \approx 0.5461 θincidentarcsin(0.5461)33.1\theta_{\text{incident}} \approx \arcsin(0.5461) \approx 33.1^\circ Answer: 33.1°

  4. Problem 4: A light ray XYXY passes from glass into a triangular wedge of air and then back into the glass. The directions of the refracted ray options are numbered 1 through 4. The correct answer based on standard refraction and reflection patterns appears to be direction 2, assuming the ray bends outward when entering air and back when re-entering glass.

Answers Recap:

  1. nliquid=1.39n_{\text{liquid}} = 1.39
  2. nglass=1.44n_{\text{glass}} = 1.44
  3. Angle of incidence 33.1\approx 33.1^\circ
  4. Correct direction: 2

Would you like further explanations or have any other questions?


Here are 5 related questions to expand on this topic:

  1. How does the refractive index affect the speed of light in a medium?
  2. Why does light bend towards the normal when entering a medium with a higher refractive index?
  3. How can total internal reflection be used in fiber optics?
  4. What happens to the angle of refraction if the refractive index of the second medium is decreased?
  5. How do you calculate the critical angle between two mediums?

Tip: Remember, the greater the refractive index, the slower light travels in that medium.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Refraction
Refractive Index
Snell's Law

Formulas

n₁sin(θ₁) = n₂sin(θ₂)
n_{liquid} = sin(θ₂) / sin(θ₁)
n_{glass} = sin(θ₁) / sin(θ₂)

Theorems

Snell's Law

Suitable Grade Level

Grade 10-12