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Chapter 30: Problem 18

A light ray propagates in a transparent material at \(15^{\circ}\) to the normal to the surface. It emerges into the surrounding air at \(24^{\circ}\) to the normal. Find the material's refractive index.

Short Answer

Expert verified
The refractive index of the material can be calculated as \(n_1 = \frac{\sin 24^{\circ}}{\sin 15^{\circ}}\).

Step by step solution

01

Understand Snell's Law

Snell's law can be stated as: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\), where \(n_1\) and \(n_2\) are the indices of refraction for the two media, and \(\theta_1\) and \(\theta_2\) are the angles the light makes with the normal to the surface. Here, the ray is moving from the material to air, thus air is the second medium with refractive index \(n_2 = 1\), because for the air, refractive index is always 1.
02

Substitute Known Values

We know that \(n_1 \sin \theta_1 = \sin \theta_2\), where \(n_1\) is the refractive index of the material, \(\theta_1\) is the angle of incidence which is 15 degrees, and \(\theta_2\) is the angle of refraction which is 24 degrees. We then substitute these values into the equation: \(n_1 \sin 15^{\circ} = \sin 24^{\circ}\).
03

Solve for the refractive index

To isolate \(n_1\) and solve for the refractive index of the material, we rearrange our equation from Step 2 to get \(n_1 = \frac{\sin 24^{\circ}}{\sin 15^{\circ}}\). This equation is simply solved by using a calculator.

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Most popular questions from this chapter

Why does a spoon appear bent when it's in a glass of water?

A scuba diver sets off a camera flash at depth \(h\) in water with refractive index \(n .\) Show that light emerges from the water's surface through a circle of diameter \(2 h / \sqrt{n^{2}-1}\)

Fermat's principle states that a light ray's path is such that the time to traverse that path is an extremum (a minimum or a maximum) when compared with times for nearby paths. Show that Fermat's principle implies Snell's law by proving that a light ray going from point \(A\) in one medium to point \(B\) in a second medium will take the least time if it obeys Snell's law.

Show that a three-dimensional corner reflector (three mutually perpendicular mirrors, or a solid cube in which total internal reflection occurs) turns an incident light ray through \(180^{\circ} .\) (Hint: Let \(\vec{q}=q_{x} \hat{\imath}+q_{y} \hat{\jmath}+q_{z} \hat{k}\) be a vector in the propagation direction. How does this vector get changed on reflection by a mirror in a plane defined by two of the coordinate axes?)

For the interface between air (refractive index 1 ) and a material with refractive index \(n\), show that the critical angle and the polarizing angle are related by \(\sin \theta_{c}=\cot \theta_{p}\)
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