3. The common musk turtle, also called a “stinkpot,” has a length of 7.60 cm at maturity. Suppose a turtle with this length is placed in front of a diverging lens that has a 14.0 cm focal length. If the turtle’s image is 4.00 cm across, how far is the turtle from the lens? How far is the turtle’s image from the lens?

1) 12.6 cm

2) -6.6 cm

Explanation:

We can solve this problem by using two equations:

1) Lens equation:

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

2) Magnification equation:

where

y' is the size of the image

y is the size of the object

In this problem we have:

y = 7.60 cm (size of the turtle)

y'= 4.00 cm (size of the image

So from eq.(2) we get

Substituting into eq(1), we can find the value of p:

Here we have

f = -14.0 cm is the focal length (negative for a diverging lens)

So we find

2)

Now we want to find how far is the turtle's image from the lens: so, we want to find the value of q.

We can do it by using the magnification equation:

where here we have:

y' = 4.00 cm is the size of the image

y = 7.60 cm is the size of the object

p = 12.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q we find:

And the negative sign means the image is virtual.

the answer is kevlar