Chapter 9: Problem 49

Two light waves that have wavelengths of 700 and $400 \mathrm{~nm}$ enter block of glass (from air) with the same angle of incidence. Which has the larger angle of refraction? Why? Would the answer be different if the light waves were going from glass into air:

Short Answer

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Explain why. Answer: The 400nm wavelength wave has a larger angle of refraction when entering a glass block from air. The reason behind this is that the refractive index of glass is higher for shorter wavelengths (400nm in this case). As a result, the shorter wavelength light bends more when entering the glass, leading to a larger angle of refraction.

Step by step solution

Snell's Law overview

Snell's Law states that for a given pair of media and a wave with a specific frequency, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. Mathematically, it can be expressed as: $$\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$$ where $i$ and $r$ are the angle of incidence and refraction, $v_1$ and $v_2$ are the respective wave velocities in the two media, and $n_1$ and $n_2$ are their refractive indices.

Calculate refractive indices for given wavelengths

For each wavelength, we need to determine the refractive index ($n_1$ for air and $n_2$ for glass). For most practical purposes, the refractive index of air is close to 1, so we will use 1 for both wavelengths. The refractive index of glass varies with the wavelength of light, and we can use the Cauchy equation to approximate it: $$n(\lambda) = A + \frac{B}{\lambda^2}$$ where $\lambda$ is the wavelength in vacuum, and $A$ and $B$ are constants that depend on the specific type of glass. We don't have the values of these constants, so we cannot proceed to calculate the refractive indices for glass. However, we can still derive the relationship between the angles of refraction and wavelengths.

Derive the relationship between angle of refraction and wavelength

From Snell's Law, we can write: $$\frac{\sin i}{\sin r_\lambda} = \frac{1}{n(\lambda)}$$ Since both wavelengths have the same angle of incidence, $$\frac{\sin i}{\sin r_{700}} = \frac{1}{n(700)}$$ and $$\frac{\sin i}{\sin r_{400}} = \frac{1}{n(400)}$$ Divide the first equation by the second: $$\frac{\sin r_{400}}{\sin r_{700}} = \frac{n(700)}{n(400)}$$ From the Cauchy equation, we know that $n(\lambda)$ decreases with increasing wavelength. Therefore, $n(700) < n(400)$.

Determine which wave has a larger angle of refraction in glass

Since $n(700) < n(400)$, the ratio $\frac{\sin r_{400}}{\sin r_{700}} > 1$. This means that $\sin r_{400} > \sin r_{700}$. Since the sine function is increasing in the range of 0 to 180 degrees, we can conclude that the angle of refraction for the 400nm wave ($r_{400}$) is larger than the angle of refraction for the 700nm wave ($r_{700}$). The reason behind this is that the refractive index of glass is higher for shorter wavelengths (400nm in this case). As a result, the shorter wavelength light bends more when entering the glass, and hence, has a larger angle of refraction.

Discuss the case where light waves go from glass to air

When the light waves go from glass to air, Snell's Law still applies: $$\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$$ In this case, $n_1$ represents the refractive index of glass (which depends on the wavelength) and $n_2$ represents the refractive index of air (which is close to 1 for both wavelengths). Using a similar analysis as before, we will still find that $r_{400} > r_{700}$, meaning that the 400nm wave has a larger angle of refraction when going from glass to air as well. The reason remains the same: higher refractive index for shorter wavelengths causes larger bending of the light wave.

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Two light waves that have wavelengths of 700 and \(400 \mathrm{~nm}\) enter block of glass (from air) with the same angle of incidence. Which has the larger angle of refraction? Why? Would the answer be different if the light waves were going from glass into air:

Short Answer

Step by step solution

Snell's Law overview

Calculate refractive indices for given wavelengths

Derive the relationship between angle of refraction and wavelength

Determine which wave has a larger angle of refraction in glass

Discuss the case where light waves go from glass to air

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