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Chapter 30: Problem 29

The refractive index of a human cornea is \(1.40 .\) If 550 -nm light strikes a cornea at incidence angle \(25^{\circ},\) find (a) the angle of refraction and (b) the wavelength in the cornea.

Short Answer

Expert verified
The angle of refraction is approximately \(17.38^{\circ}\) and the wavelength in the cornea is approximately \(392.86 nm\).

Step by step solution

01

Apply Snell's Law to find the angle of refraction

Snell's law states that \(n_1 \cdot sin(\Theta_1) = n_2 \cdot sin(\Theta_2)\) where \(\Theta_1\) and \(\Theta_2\) are the incidence and refraction angles respectively, and \(n_1\) and \(n_2\) are the refractive indices. Here, \(n_1\) is the refractive index of air, which is 1, and \(\Theta_1\) is 25 degrees. \(n_2\) is the refractive index of the cornea, which is 1.40. So, rearranging for \(\Theta_2\), we have \(\Theta_2 = arcsin[\frac{n_1 \cdot sin(\Theta_1)}{n_2}]\).
02

Calculate the angle of refraction

Substituting given values into the formula from Step 1, we get \(\Theta_2 = arcsin[\frac{1 \cdot sin(25^{\circ})}{1.40}]\). After calculating, we get \(\Theta_2 \approx 17.38^{\circ}\).
03

Calculate the wavelength in the cornea

The wavelength of light in a medium is given by \(\lambda_2 = \frac{\lambda_1}{n_2}\), where \(n_2\) is the refractive index of the medium, and \(\lambda_1\) is the wavelength in vacuum or air. Here, \(\lambda_1 = 550\) nm and \(n_2 = 1.40\). So, \(\lambda_2 = \frac{550 nm}{1.40}\). After calculating, we get \(\lambda_2 \approx 392.86 nm\).

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