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Chapter 23: Problem 15

A laser beam in air is incident on a liquid at an angle of \(37^{\circ}\) with respect to the normal. The laser beam's angle in the liquid is \(26^{\circ} .\) What is the liquid's index of refraction?

Short Answer

Expert verified
The index of refraction of the liquid is approximately 1.34, when rounding to two decimal places.

Step by step solution

01

Identifying Known Variables

Identify the known variables from the problem. Here, the angle of incidence in air (\(i\)) is \(37^{\circ}\) and the angle of refraction in the liquid (\(r\)) is \(26^{\circ} .\) In addition, it's known that the refractive index of air (\(n_1\)) is approximately equal to 1.
02

Applying Snell's Law

Snell's law is given by the formula \( n_1 \cdot \sin(i) = n_2 \cdot \sin(r)\) where \(n_1\) and \(n_2\) are the indices of refraction of the two media, i and r are the angles of incidence and refraction respectively. Here we want to solve for \(n_2\), so rearrange the formula and replace other variables with the known quantities: \(n_2 = \frac{n_1 \cdot \sin(i)}{\sin(r)}.\)
03

Calculating the result

Substitute the known values into the adjusted formula: \( n_2 = \frac{1 \cdot \sin(37^{\circ})}{\sin(26^{\circ})}\) Perform the calculation to find the index of refraction for the liquid.

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