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.
