Shown below is a ray of light going from air through crown glass into water, such as going into a fish tank. Calculate the amount the ray is displaced by the glass \((\Delta x), \quad\) given that the incident angle is \(40.0^{\circ}\) and the glass is \(1.00 \mathrm{cm}\) thick.

Short Answer

Expert verified
To find the displacement of the light ray caused by the glass, first use Snell's Law to find the angle of refraction in the glass: \(\theta _2 = \arcsin \frac{1.0003 \cdot \sin 40^{\circ}}{1.52}\). Then calculate the path length inside the glass: \(L = \frac{1.00 \,\text{cm}}{\cos \theta _2}\) and the angle of refraction in water: \(\theta _3 = \arcsin \frac{1.52 \cdot \sin \theta _2}{1.333}\). Finally, find the horizontal distance traveled by the light ray in the glass (x) and the horizontal distance the light ray would have traveled in a straight line without the glass (y): \(x = L \cdot \sin \theta _2\) and \(y = d \tan 40^{\circ}\). The displacement \(\Delta x = x - y\).

Step by step solution

01

Find the angle of refraction in the glass using Snell's Law

Snell's Law relates the angles of incidence and refraction to the refractive indices of the materials. The law is written as: \( n_1 \sin \theta _1 = n_2 \sin \theta _2 \) where \(n_1\) and \(n_2\) are the refractive indices of the materials, and \(\theta _1\) and \(\theta _2\) are the angles of incidence and refraction, respectively. The refractive indices for air, crown glass, and water are approximately 1.0003, 1.52, and 1.333, respectively. Given the incident angle of \(40^{\circ}\), we can find the angle of refraction in the glass (\(\theta _2\)): \( n_\text{air} \sin \theta _1 = n_\text{glass} \sin \theta _2 \) \( 1.0003 \cdot \sin 40^{\circ} = 1.52 \cdot \sin \theta _2 \) Solve for \(\theta _2\): \( \theta _2 = \arcsin \frac{1.0003 \cdot \sin 40^{\circ}}{1.52}\)
02

Calculate the path length of the light ray inside the glass

Now find the length of the light ray inside the glass. Use the thickness of the glass (d) and the angle of refraction to calculate the path length (L) using basic trigonometry: \( L = \frac{d}{\cos \theta _2} \) where d is the thickness of the glass (1.00 cm). Substitute the thickness and the angle of refraction: \( L = \frac{1.00 \,\text{cm}}{\cos \theta _2} \)
03

Calculate the angle of refraction in the water

Find the angle of refraction in water (\(\theta _3\)), again using Snell's Law: \( n_\text{glass} \sin \theta _2 = n_\text{water} \sin \theta _3 \) \( 1.52 \cdot \sin \theta _2 = 1.333 \cdot \sin \theta _3 \) Solve for \(\theta _3\): \( \theta _3 = \arcsin \frac{1.52 \cdot \sin \theta _2}{1.333}\)
04

Calculate the displacement of the light ray

Calculate the horizontal distance traveled by the light ray in the glass (x) using trigonometry: \( x = L \cdot \sin \theta _2 \) Calculate the horizontal distance the light ray would have traveled in a straight line without the glass (y): \( y = d \tan 40^{\circ} \) Finally, find the displacement of the light ray by subtracting y from x: \( \Delta x = x - y \)

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