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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 \(\theta_2\) and the wavelength in the cornea \(\lambda_{cornea}\) can be found using Snell's law and frequency conservation law.

Step by step solution

01

Calculation of the angle of refraction

Firstly, apply Snell's Law to calculate the angle of refraction. The problem provides the refractive index of the cornea \(n_2=1.40\), the angle of incidence \(\theta_1= 25^{\circ}\) and the refractive index of air \(n_1=1\). The angle of refraction \(\theta_2\) can be found using the equation \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). Therefore, \(\theta_2 = \arcsin\left(\dfrac{n_1}{n_2} \sin \theta_1 \right)\).
02

Calculation of the wavelength in the cornea

When light goes from one medium to another, the frequency stays constant but the speed and wavelength of light changes. Specifically, the wavelength changes with speed while frequency stays constant. The wavelength of light in a medium with refractive index \(n\) is given by \(\lambda_n = \dfrac{\lambda_0}{n}\), where \(\lambda_0\) is the wavelength in vacuum and \(n\) is refractive index of medium. Here \(\lambda_0 = 550 \, nm\), so we can find the wavelength in cornea \(\lambda_{cornea} = \dfrac{\lambda_0}{n_2}\).

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