Two students in a physics laboratory each have a concave mirror with the same radius of curvature, 50 cm. each student places an object in front of a mirror. the image in both mirrors is three times the size of the object. however, when the students compare notes, they find that the object distances are not the same. what is the distance of the closer object?

16.67 cm.

Explanation:

radius of curvature r = 50 cm

Focal length f = 25 cm

Magnification can be achieved 3 times in concave mirror in two ways .

1) By putting the object very close to the mirror ( within focal length ) so that magnified virtual and erect image is observed.

2) By putting the object between f and 2f so that real and inverted magnified image is observed .

In the question , the object distance in the former case has been asked.

\frac{v}{u}=3\\

v = 3u

Using the mirror formula we can calculate the required distance as follows

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Put v = -3u, f = -25.

-\frac{1}{3u}+\frac{1}{u}=\frac{1}{-25}\\

u=16.67cm

your answer is c! hope this : )

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .