A student wants to obtain the image of an object place at 20cm from the pole of a mirror. If the screen is placed at a distance of 50cm find the focal length of the mirror name the mirror and nature and size of image so formed
Answers
Ray diagrams can be used to determine the image location, size, orientation and type of image formed of
objects when placed at a given location in front of a concave mirror. The use of these diagrams was demonstrated earlier in Lesson 3. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size.
To obtain this type of numerical information, it is necessary to use the
Mirror Equation
and the
Magnification Equation
. The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
The magnification equation relates the ratio of the image distance and object-distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.
ref.The Mirror Equation
In the present problem, the object is placed at a position which is between f and 2f so the image will be formed between infinity and 2f.
logic is had the object been at f …the image will be at infinity and if it is at 2f the image will come to 2f
using the mirror relation
1/v + 1/u = 1/f where u, v, f are object distance, image distance, and focal length respectively.
u= 20 cm f=15 cm (the sign convention is a coordinate convention with the mirror at origin and object on the positive x-axis)
so 1/v = 1/f - 1/u , = 1/15 - 1/20
= 5/(15x20)= 1/60 therefore v= 60 cm
so the image will be formed at 60 cm
with magnification m= v/u = 60/20 = 3
the image will be 3 times larger and real