You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with a focal length of 6.0 cm and focal length of 12.0 cm. For both lenses, the light fills the size of the lens. Using the Gaussian beam equations, what is the smallest diameter of the beam (known as the beam waist) for each lens

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = = -

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .