Para los lentes, tenemos unos en los cuales, las imagenes son reales o virtuales, mas grandes o mas pequeñas, derechas o invertidas. Un lapiz de 15cm de altura se coloca a 6.0cm a la izquierda de una lente divergente y se obtiene una imagen de el a 2.0cm a la izquierda del lente.
a. Encuentre la distancia focal de este lente.
b. Encuentre la altura de la imagen.
The lens given is a diverging lens. The object distance from the lens is 6cm to the left of the lens, the height of the object is ho = 15cm and and the image distance is 2cm to the left of the lens.
By sign convention, the object distance u = -6cm and the image distance v = -2cm
a) The lens equation is given by,
So the focal length of the lens is 3cm. The negative sign shows that it is a diverging lens.
b) The magnification of thelens is given by,
So the height of the image is 5cm.
The image is virtual, straight and smaller than the object.
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