The equation of the constructor of the geomeric optics allows to find the results for the distance and size of the objects are:
The equation of the constructor of geometric optics describes the position of objects and their images for mirrors and lenses.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distances to the object and the image, respectively.
They indicate the focal length f = 10 cm, the distance to the image q = 40 cm, that the height s h ’= 4 cm. In the attachment we see a scheme of the system.
Let's find the distance to the object.
[tex]\frac{1}{p\frac{x}{y} } = \frac{1}{f} - \frac{1}{q} \\\frac{1}{p} = \frac{1}{10} - \frac{1}{40}\\\frac{1}{p} = 0.075[/tex]
p = 13.3 cm
The magnification in the ratio of image size to object size. The image is inverted so its height is negative.
[tex]m = \frac{h'}{h} = - \frac{q}{p}\\h = - \frac{p}{q} \ h'\\h = - \frac{13.3}{40}\ (-4)[/tex]
h = 1.33 cm
In conclusion using the equation of the constructor we can find the results for the distance and size of the objects are:
Learn more here: brainly.com/question/14314471
