Physics Optics Help

Question

An object that is 45cm away from a lens has a magnification of 2.5.

1. What is the focal distance of this lens?

Is the image created real or virtual and how do you know?

Answer 1

The two equations that we will be working with are

M = -di/do (magnification = -image distance / object distance)

and

1/f = 1/do + 1/di where f is the focal length (what we are looking for)

The most important piece of conceptual information here is that the magnification is greater than 1. This tells us right away that the image is virtual!

Use M equation to find di

M=2.5 do= 45cm di= ?

2.5 = - di / 45 multiply by 45 on both sides

112.5 = -di divide by -1 on both sides

-112.5 cm = di

di should be negative if it is showing up on the same side of the lens as do, which confirms that it is not a real image.

Lastly, plug into focal length equation to find f

1/f = 1/do + 1/di

Remember, to create a virtual image, the object must be in front of the focal point, so we expect our answer for f to be more than 45cm. (This is where the conceptual comes in handy - plug in your value of di in such a way that makes this true —> this will show why di should be negative).

1/f = 1/45 + -1/112.5 add the fractions together using a calculator or by common denominator

Common denominator is 225

1/45 —> 5/225

-1/112.5 —> -2/225

5/225 - 2/225 = 3/225

or if done by calculator = 0.01333

1/f = 3/225 or 1/f = 0.01333 invert by flipping numerator and denominator (or using x^-1 or 1/x button on calculator)

f/1 = 225/3 or f = 0.01333^-1

F = 75cm

We know it’s a virtual image because

1) f was greater than do (object was in front of the focal point)

2) M was greater than 1

3) di was negative when do was positive