# Light Refraction Problem - Snells Law

In summary, the conversation discusses a scenario where a room key is lost at the bottom of a pool and needs to be retrieved. The method used involves using a flashlight and the law of sines to calculate the distance of the key from the edge of the pool. The final answer is determined to be 4.4m.

## Homework Statement

After swimming, you realize you lost your room key whilst in the pool. You go back at night and use a flashlight to try and find it. The flashlight shines on it, which is at the bottom of the pool, when the flashlight is held 1.2m above the surface and is directed towards a point on the surface which is 1.5m from the edge of the pool. If the pool is 4m deep, how far is the key from the edge of the pool? Take the refractive index of the water to be 1.33.

Diagram: http://oi58.tinypic.com/2s1qfqp.jpg

## Homework Equations

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$
Law of Sines

## The Attempt at a Solution

Well, what I did first was trying to find out the angle of θ_1 , the angle denoted θ_i in the diagram.

To do that I used the law of sines and the fact that boths the angles are encompassed by right angled trianlges. First I found the length of the hypontenuse (the line of sight of the flashlight)

$l=\sqrt{1.5^2+1.2^2}=1.921$ , then with that length use the law of sines

$\frac{sin90}{1.921}=\frac{\theta_1}{1.5}=0.5206 \\ \therefore \theta_1=sin^{-1}(1.5 \times 0.5206) = 51.343°$

Now used snells law to find θ_2 ...

$\frac{sin51.343}{sin\theta_2}=\frac{1.33}{1}=1.33 \\ sin\theta_2=\frac{sin51.343}{1.33}=0.587 \\ \therefore \theta_2=sin^{-1}0.587=35.94°$

Then used the law of sines again to find the value for x on the diagram

$\frac{sin(180-90-35.94)}{4}=\frac{sin35.94}{x} \\ \frac{sin54.06}{4}=\frac{sin35.94}{x} \\ 0.2024=\frac{sin35.94}{x} \\ \therefore x=\frac{sin35.94}{0.2024}=2.9m$

And therefore the answer would be 2.9+1.5=4.4m

Im sure there must be loads of rounding errors in this, but I used 4 decimal places to try and minimize them. And there is probably easier ways to go about it, but would really appreciate it if someone could see if I have done it correctly please?

Thanks :)

Last edited:
Excellent work. Clear picture, good post. I get the same answer. No easier way.

If you really insist on improving something: for the calculation of the result, you need ##\sin\theta_1## and all you have is ##\tan\theta_1 = 1.5/1.2##, so ##\sin (\tan^{-1}(1.5/1.2) ) ## is what I would calculate. Not really different or easier. Law of sines is OK, but you want to pick out sines, cosines, tangents more or less directly without much ado.

BvU said:
Excellent work. Clear picture, good post. I get the same answer. No easier way.

If you really insist on improving something: for the calculation of the result, you need ##\sin\theta_1## and all you have is ##\tan\theta_1 = 1.5/1.2##, so ##\sin (\tan^{-1}(1.5/1.2) ) ## is what I would calculate. Not really different or easier. Law of sines is OK, but you want to pick out sines, cosines, tangents more or less directly without much ado.

OK great to know it was correct. Thanks :)

## 1. What is Snell's Law?

Snell's Law, also known as the law of refraction, is a formula that describes the relationship between the angle of incidence and the angle of refraction when light passes through different mediums, such as air, water, or glass.

## 2. How does light refract when passing through a medium?

When light passes through a medium, it changes speed and bends due to the change in density. This bending of light is known as refraction. The amount of bending is determined by the angle of incidence, the angle of refraction, and the refractive index of the medium.

## 3. What factors affect the angle of refraction?

The angle of refraction is affected by the change in speed of light as it passes through different mediums, as well as the angle of incidence. The refractive index, which is a measure of how much a medium can slow down light, also plays a significant role in determining the angle of refraction.

## 4. How is Snell's Law used in everyday life?

Snell's Law is used in everyday life in various ways, such as the design of eyeglasses and contact lenses to correct vision, the creation of optical instruments like microscopes and telescopes, and the study of atmospheric refraction to understand how light travels through the Earth's atmosphere.

## 5. What are some applications of Snell's Law in scientific research?

The study of Snell's Law has been crucial in various scientific fields, such as optics, materials science, and geophysics. It has been used to develop advanced imaging techniques, understand the properties of different materials, and analyze the behavior of seismic waves in the Earth's crust.

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