Problem 33 - University Physics with Modern Physics

Question

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portions back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 \(\mu\)m in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34. (a) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? (b) What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

Step-by-Step Solution

Answer

(a) 0.375 J; (b) ~5.12 x 10^-7 Pa; (c) 604.48 nm, 3.70 x 10^14 Hz; (d) 677 V/m, 2.26 x 10^-6 T.

Step 1: Calculating Energy per Pulse

The energy delivered by each pulse is given by the formula \( E = P \times t \), where \( P \) is the power of the laser (250 mW) and \( t \) is the pulse duration (1.50 ms). Convert 250 mW to watts and 1.50 ms to seconds before calculating: \( E = 0.250 \text{ W} \times 0.00150 \text{ s} = 0.000375 \text{ J} \). So, the energy per pulse is 0.375 J.

Step 2: Calculating Average Pressure on Surface

The average pressure exerted by the laser beam when fully absorbed is given by \( p = \frac{F}{A} = \frac{P}{A} \times c \), where \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\)). Use \( A = \pi \left(\frac{d}{2}\right)^2 \) to find the area, with \( d \) given as 510 \(\mu\text{m}\) or 0.000510 m. Thus, \[ A = \pi \left(\frac{0.000510}{2}\right)^2 \approx 2.045 \times 10^{-7} \text{ m}^2 \]. \( F/P = \text{Intensity} = P/A \) and pressure \( p = (P/A) \times \frac{1}{c} \approx 5.12 \times 10^{-7} \text{ Pa} \).

Step 3: Calculating Wavelength and Frequency in Vitreous Humor

Inside the vitreous humor, the wavelength \( \lambda_{\text{medium}} \) changes due to the medium's refractive index \( n = 1.34 \). Use the formula \( \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \) where \( \lambda_{\text{vacuum}} = 810 \text{ nm} \). Thus, \[ \lambda_{\text{medium}} = \frac{810}{1.34} \approx 604.48 \text{ nm} \]. The frequency \( f \) remains unchanged, \( f = \frac{c}{\lambda_{\text{vacuum}}} \approx 3.70 \times 10^{14} \text{ Hz} \), as frequency doesn't change from air to medium.

Step 4: Finding Maximum Electric and Magnetic Fields

The intensity of light relates to the electric field \( E_{\text{max}} \) using \( I = \frac{1}{2} \varepsilon_0 c E_{\max}^2 \), where \( \varepsilon_0 \) is the vacuum permittivity and \( I = \frac{P}{A} \approx 611 \text{ W/m}^2 \). Solve for \( E_{\text{max}} \). The magnetic field \( B_{\text{max}} = \frac{E_{\text{max}}}{c} \) comes from \( c E_{\text{max}} = B_{\text{max}} \). Assuming calculations, \( E_{\text{max}} \approx 677 \text{ V/m} \) and \( B_{\text{max}} \approx 2.26 \times 10^{-6} \text{ T} \).

Key Concepts

Retina Repair with Laser Technology

Lasers are pivotal in modern medical procedures, including retina repair. The retina is the light-sensitive layer at the back of the eye, crucial for vision. When parts of the retina detach, vision impairment occurs. Laser therapy is used to repair this by "welding" the detached portion back into place. Here's how it works:

• A high-intensity laser beam is directed at the detached retina.
• This beam is typically short and powerful, ensuring precision and minimizing damage to surrounding tissue.
• The energy from the laser creates tiny burns, causing scar tissue that reattaches the retina.

The procedure is efficient, often completed in an outpatient setting. It's a safe alternative to more invasive surgeries. Understanding how laser beams work in this context highlights their importance in modern medical solutions.

Energy Calculation of Laser Pulses

Calculating the energy delivered by a laser pulse is essential for understanding its impact. In the given exercise, the laser emits pulses of energy, each lasting 1.50 milliseconds with a power of 250 mW.

• First, convert power from milliwatts to watts: 250 mW equals 0.250 watts.
• Convert time from milliseconds to seconds: 1.50 ms equals 0.00150 seconds.
• Multiply these to find energy: Energy \( E = P \times t = 0.250 \text{ W} \times 0.00150 \text{ s} = 0.000375 \text{ J} \).

This calculation shows that each laser pulse delivers 0.375 joules of energy to the retina. Knowing this energy output helps ensure the beam's effectiveness and safety, vital for both patient outcomes and equipment specifications.

Understanding Index of Refraction

The index of refraction is a measure of how much a medium, like the vitreous humor in the eye, slows down the speed of light passing through it. The index is defined as:

• A ratio of the speed of light in a vacuum to its speed in the medium.
• The formula is \( n = \frac{c}{v} \), where \( n \) is the index of refraction, \( c \) is the speed of light in vacuum, and \( v \) is the speed in the medium.
• The vitreous humor has an index of 1.34, indicating it slows light more than air does.

The higher the index, the more it bends light towards the normal line when entering the medium. This understanding is crucial when working with optics and lasers, as it affects how light behaves in mediums, ensuring accurate focus and alignment during procedures like retina repair.

Wavelength and Frequency in Mediums

When light enters a medium with an index of refraction different from that of a vacuum, its wavelength changes while its frequency remains constant. In the exercise, an 810 nm laser wavelength changes when entering the vitreous humor.

• The new wavelength is calculated using \( \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \), where \( n = 1.34 \).
• Thus, \( \lambda_{\text{medium}} = \frac{810 \text{ nm}}{1.34} \approx 604.48 \text{ nm} \).
• Frequency \( f \) remains unchanged as frequency is unaffected by the medium: \( f = \frac{c}{\lambda_{\text{vacuum}}} \approx 3.70 \times 10^{14} \text{ Hz} \).

This principle is vital in laser applications, ensuring the desired effect of the laser pulse is maintained even when entering different mediums. It guarantees that expectations for energy delivery and interaction within the eye remain consistent.