Question

P3.4. The spot size is clearly extremely important in determining the power density of a focused laser beam, an important determi- nant of how it may be used in surgery. However, there is a basic limitation on how small a spot one can achieve with a given wavelength laser beam, ; a given diameter laser beam (D); and a lens with focal length, f. This smallest spot sizc is given by: where d is the diameter of the focused laser beam at the focal point of the lens (Figure P3.4). This cquation can be derived using the wave theory of light, but we will not examine the mathemat ics of this derivation Use the above formula for the spot size, d, and a wavelength of 514.5 nm for an argon laser, to estimate the spot size of the laser inside the cye after the laser has passed through the lens of the eyc. The cye's focal length is approximately f 2 cm (the approximate distance from the lens to the retina). Explain why ophthalmological surgeons would use a lens to defocus (sprcad out) the laser beam before it enters the cyc a. when performing photocoagulation b. Find an equation for the ratio of the power densities of two lasers with different wavelengths, and2, but with P, and the same initial beam diamcter, D Assume the power density is given by I P/id/2)2. Assume each laser beam has been focused by a lens with the same focal length, f. (In fact, most laser beams used in medicine start out with a diameter close to D 1 mm, so this is a real istic assumption.) Which laser beam is focused to the higher power spot-the one with the shorter or the one with the the samc power, an longer wavelength? c. Next, use your answer from part b to compare quantita t result in the follow- tively the different power densitics t ing cases: i. The three different wavelengths at which the Nd:YAG laser operates (1065 nm, 532 nm, or 355 nm; assume the samc powcr is emitted at cach wavclength)

Answer 1

a)

21 f

given :

514.5 nm = 515.5 × 10-cm λ

  

f= 2 cm

therefore the spot size is given by substituting these values into equation (1)

2 × 514.5 × 10 × 2

(In the question D is not explicitly given, we shall assume a standard medical argon laser with has a initial diameter D = 1 mm)

2 × 514.5 × 10 × 2 = 6.55 × 10-4 crn π 1 × 10-1

The laser beam is made to spread (i.e increasing the size of initial beam D) in order to reduce the intensity and hence the power of the beam so as not to affect the sensitive retina.

b) To find ratio of Power density

the Power density is given by:

$I = \frac{P}{\pi (\frac{d}{2})^{2}} \ \ \ \ \ \ \ \ \ \ ...(2)$

Given that P is the same for both lasers and the initial diameter D is the same too we need to find

determining I₁

(3) li =AP

so we need d₁ which we get by substituting $\lambda = {\lambda}_1$ into equation (1)

$d_1 = \frac{2 \ \lambda_1 \ f}{\pi \ D} \ \ \ \ \ \ \ \ \ \ ...(4)$

therefore equation (3) now reads

$I_1 = \frac{P}{\pi (\frac{\frac{2 \ \lambda_1 \ f}{\pi \ D} }{2})^{2}} \ \ \ \ \ \ \ \ \ \ ...(5)$ simplifying we get $I_1 = \frac{\pi \ D^2 \ P}{\lambda_1^2 \ f^2} \ \ \ \ \ \ \ \ \ \ ...(6)$

determining I₂

In a similar way we can get

$I_2 = \frac{P}{\pi (\frac{\frac{2 \ \lambda_2 \ f}{\pi \ D} }{2})^{2}}$ on simplifying we get $I_2 = \frac{\pi \ D^2 \ P}{\lambda_2^2 \ f^2} \ \ \ \ \ \ \ \ \ \ ...(7)$

therefore

will give us $\frac{1}{\lambda_1^2}:\frac{1}{\lambda_2^2} \ \ \ \ \ \ \ \ \ \ ...(8)$

therefore we see that the power density is inversely proportional to the wavelength. Therefore the laser beam with the lower wavelength should be focused to the spot that needs higher power.

c)

Using equation (8) and expanding for three variables we have:

is  $\frac{1}{\lambda_1^2}:\frac{1}{\lambda_2^2} :\frac{1}{\lambda_3^2}\ \ \ \ \ \ \ \ \ \ ...(9)$

therefore

$\frac{1}{\lambda_1^2}:\frac{1}{\lambda_2^2} :\frac{1}{\lambda_3^2} =\frac{1}{1065^2}:\frac{1}{532^2} :\frac{1}{355^2}$