PROBLEMAS DE ESTUDIO DE ESPEJOS
1. a) We know that f = 20.0 cm for this particular thin lens. If the object is
placed at a distance of p = 40.0 cm, then
1/(40.0 cm) + 1/q = 1/(20.0 cm)
1/q = 1/(20.0 cm) - 1/(40.0 cm) = 1/(40.0 cm)
Thus, q = 40.0 cm. If we use this in the magnification equation, we get M
= -q/p = 1(40.0 cm)/(40.0 cm) = -1. Hence, the image is real, inverted, the
same size, and a distance of 40.0 cm behind the lens.
b) For p = 20.0 cm,
1/(20.0 cm) + 1/q = 1/(20.0 cm)
1/q = 0
Thus , q is equal to infinity. This means that no image is formed, and the
rays are parallel.
c) For p = 10.0 cm,
1/(10.0 cm) + 1/q = 1/(20.0 cm)
1/q = 1/(20.0 cm) - 1/(10.0 cm) = -1/(20.0 cm)
Thus, q = -20.0 cm. For the magnification, we get M = - q/p = -(-20.0
cm)/(10.0 cm) = 2. Hence, the image is upright, magnified by 2, and 20.0
cm in front of the lens.
INTERFERENCIA DE LA LUZ
DIFRACCION Y POLARIZACION
Wavelength λ = 589 nm
= 589 * 10^–9 m
Aperture of the objective = D = 0.90 cm
= 0.90 * 10^–2 m
a)
αmin = 1.22*λ/D
= 1.22*589*10^-9/0.9*10^-2
= 8*10^-5 rad.
b)
For maximum limit of resolution the shortest wavelength of the visible
spectrum is used which is 400 nm for violet colour light.
Using the formula
αmin = 1.22*λ/D
= 1.22*400*10^-9/0.9*10^-2
= 5.4*10^-5 rad.
0
