Anonymous
Anonymous asked in 科學及數學其他 - 科學 · 1 decade ago

Lens!!!!幫幫手please!!!!!!!

1) An object is placed in front of a converging lens. An image is formed 30 cm in front of the lens. The focal length of the lens is 20cm.

A. Draw the ray diagram.

B. State the nature of the image.

.

我想問點搵個object/image幾高? 定係自己let佢幾多就幾多?

同埋個ans話個image係real?乜唔係virtual咩?點知佢係real/virtual image?

2) An object located in front of a diverging lens. An image is formed 30 cm from the lens. The focal length of the lens is 50cm

a)Draw the ray diagram

咩係from the lens 呀...? 係個 lens右邊?

要點畫ar?pleasehelp...

Update:

can you add some pictures to describe?

anyway, I will choose you to be the best one who answers the question later

1 Answer

Rating
  • 天同
    Lv 7
    1 decade ago
    Favorite Answer

    1. Since the ratio of the height of image to the height of object for a given object distance from the lens is a constant (this is the magnification), you could use an arbitrary height for the object/image.

    Since the image is formed by a convex lens at the same side as the object, it must be virtual and erect.

    For a virtual image, the light rays do not actually intersect at the image position, and only be seen as if they come from the image.

    Hence, adraw an erected arrow on the principal axis at 30 cm from the lens to represent the image. At its tip, draw a straight ray passing the lens centre (this ray direction is unaffected by the lens). Then draw another straight ray from the tip of the image to the lens focus at the other side of the lens. At the point where this second ray meets the lens, construct a line parallel to the principal axis and which will cut the first light ray at a point. This point will be the position of the tip of the object.

    Finally, re-draw the light rays nicely that originate from the object.

    ---------------------------------------------------

    2. It is given that the lens is a divergent lens (concave lens), images formed by concave lens are ALWAYS VIRTUAL. Therefore, it must be on the same side of the lens as the object, and is erect.

    You could use a method similar to that given in Q1 to locate the object and then the correct ray diagram. In this case, you have to make use of the focus at the same side of the lens where the object and image are located.

Still have questions? Get your answers by asking now.