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Chapter 30: Problem 26

Blue and red laser beams strike an air-glass interface with incidence angle \(50^{\circ} .\) If the glass has refractive indices of 1.680 for the blue light and 1.621 for the red, what will be the angle between the two beams in the glass?

Short Answer

Expert verified
The angle between the blue and red light beams in the glass, \( \theta_{br} \), is the difference between \( \theta_{b} \) and \( \theta_{r} \)

Step by step solution

01

Apply Snell's Law for Blue Light

Snell's law can be written as \( n_1\sin(\theta_1) = n_2\sin(\theta_2) \) where \( n_1, n_2 \) are the refractive indices of the materials, and \( \theta_1, \theta_2 \) are the angles made with the normal. Apply this for blue light. The refractive index of air is approximately 1, so \( 1\sin(50^{\circ}) = 1.680\sin(\theta_{b}) \) where \( \theta_{b} \) is the angle that the blue light makes in the glass.
02

Solve for the Blue Light Angle

To get the blue light angle, rearrange the equation and solve for \( \theta_{b} \), which results to \( \theta_{b} = \sin^{-1}\left(\frac{1\sin(50^{\circ})}{1.680}\right) \) .
03

Apply Snell's Law for Red Light

Repeat steps 1 and 2 for the red light. The refraction equation becomes \( 1\sin(50^{\circ}) = 1.621\sin(\theta_{r}) \) , where \( \theta_{r} \) is the angle for the red light in the glass.
04

Solve for the Red Light Angle

Solving for \( \theta_{r} \) becomes \( \theta_{r} = \sin^{-1}\left(\frac{1\sin(50^{\circ})}{1.621}\right) \) .
05

Find the Angle Between the Beams

Finally, find the angle between the beams by subtracting one beam's angle from the other. \( \theta_{br} = \theta_{b} - \theta_{r} \)

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