Question

wavelength, frequency, and the speed of light in different media
a beam of light from a monochromatic laser shines into a piece of glass. the glass has thickness l and index of refraction n = 1.5. the wavelength of the laser light in vacuum is l/10 and its frequency is f. in this problem, neither the constant c nor its numerical value should appear in any of your answers.
part a
how long does it take for a short pulse of light to travel from one end of the glass to the other?
express your answer in terms of the frequency, f. use the numeric value given for n in the introduction.
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Explanation:

Step1: Recall speed in medium

The speed of light in a medium with refractive index \( n \) is \( v=\frac{c}{n} \), but we can also relate speed to wavelength and frequency. In vacuum, \( c = \lambda_0 f \), where \( \lambda_0=\frac{L}{10} \) (given). In the glass, the wavelength \( \lambda=\frac{\lambda_0}{n} \) (since wavelength in medium is vacuum wavelength divided by \( n \)), and the frequency \( f \) remains the same (frequency is invariant when light enters a medium). So the speed in glass \( v = \lambda f=\frac{\lambda_0}{n}f \). Substituting \( \lambda_0=\frac{L}{10} \), we get \( v=\frac{Lf}{10n} \).

Step2: Calculate time

Time \( t \) is distance \( L \) divided by speed \( v \), so \( t = \frac{L}{v} \). Substituting \( v=\frac{Lf}{10n} \) into this, we have \( t=\frac{L}{\frac{Lf}{10n}}=\frac{10n}{f} \). Given \( n = 1.5 \), so \( t=\frac{10\times1.5}{f}=\frac{15}{f} \). Wait, wait, no—wait, let's re - do. Wait, the wavelength in vacuum is \( \lambda_0=\frac{L}{10} \), and \( c=\lambda_0 f \). In the glass, speed \( v=\frac{c}{n}=\frac{\lambda_0 f}{n} \). Then time \( t=\frac{L}{v}=\frac{L n}{\lambda_0 f} \). Substitute \( \lambda_0=\frac{L}{10} \), so \( t=\frac{L n}{\frac{L}{10}f}=\frac{10n}{f} \). With \( n = 1.5 \), \( 10n = 15 \), so \( t=\frac{15}{f} \)? Wait, no, wait, maybe I messed up the wavelength. Wait, the problem says "the wavelength of the laser light in vacuum is \( L/10 \)". So \( \lambda_0=\frac{L}{10} \), and \( c=\lambda_0 f \). In the glass, the speed is \( v=\frac{c}{n}=\frac{\lambda_0 f}{n} \). The distance is \( L \), so time \( t=\frac{L}{v}=\frac{L n}{\lambda_0 f} \). Substitute \( \lambda_0 = \frac{L}{10} \), so \( t=\frac{L n}{\frac{L}{10}f}=\frac{10n}{f} \). Now, \( n = 1.5 \), so \( 10\times1.5 = 15 \), so \( t=\frac{15}{f} \)? Wait, no, wait, let's check again. Wait, the thickness is \( L \), speed in glass is \( v=\frac{c}{n} \), and \( c=\lambda_0 f \), \( \lambda_0=\frac{L}{10} \), so \( c=\frac{Lf}{10} \), so \( v=\frac{Lf}{10n} \). Then time \( t=\frac{L}{v}=\frac{L}{\frac{Lf}{10n}}=\frac{10n}{f} \). Yes, that's correct. So with \( n = 1.5 \), \( 10n=15 \), so \( t = \frac{15}{f} \)? Wait, no, wait, 10 times 1.5 is 15? Wait, 10*1.5 = 15, yes. So \( t=\frac{15}{f} \)? Wait, but let's think again. Alternatively, since \( c=\lambda_0 f \), \( \lambda_0 = L/10 \), so \( c=(L/10)f \). Speed in glass \( v = c/n=(Lf)/(10n) \). Time \( t = L/v = L/(Lf/(10n))=10n/f \). Plugging \( n = 1.5 \), we get \( t=(10\times1.5)/f = 15/f \). Wait, but maybe the problem is simpler. Wait, the key is that frequency \( f \) is related to wavelength in vacuum \( \lambda_0 \) by \( c=\lambda_0 f \). In the glass, the speed is \( v = c/n \), so \( v=\lambda_0 f/n \). The distance is \( L \), so time \( t = L/v = Ln/(\lambda_0 f) \). Since \( \lambda_0 = L/10 \), substitute: \( t = Ln/( (L/10)f )=10n/f \). With \( n = 1.5 \), \( 10n = 15 \), so \( t = 15/f \). Wait, but let's check units. Wavelength \( \lambda_0 \) has units of length, \( f \) is frequency (1/time), so \( c=\lambda_0 f \) has units of length/time, correct. Speed in glass \( v = c/n \), length/time. Time \( t = L/v \), length/(length/time)=time, correct. So the calculation seems right.

Answer:

\( \frac{15}{f} \) (or \( \frac{10\times1.5}{f} \), but simplified as \( \frac{15}{f} \))