1. A real object 200 mm away from a 100 mm positive lens is imaged where? …. And with what magnification?

Solution: I will use the lens maker's equation in the Cartesian coordinate system:

1/s' = (1/F) + (1/s)

Since the object is assumed to be real and to the left of the lens, it's distance is negative

s' = [(1/F) + (1/s)]^{-1} = [(1/100) + (1/{-200})]^{-1 }= [(1/100) - (1/200)]^{-1}

s' = + 200 mm, so this is a real image to the right of the lens

What is the magnification? M = s'/s = -200/200

M = -1.0

This is referred to as a the 2F to 2F imaging system... both object and image are 2F away and it provides unity magnification.

2. A virtual object is 30 mm inside a negative lens of focal length -100 mm. Where is the image?

Warning: this is intended to tax your understanding of imaging and it is a *difficult* problem. I include this to motivate the student to embrace the imaging nomograph I discuss later:)

Solution: The virtual object means there is another optical system to the left of this lens and it is putting an image to the right of the lens. Said another way the object is imaged inside the lens! In this case the object distance is positive! Using the lens maker's equation in the Cartesian coordinate system:

1/s' = (1/F) + (1/s)

s' = [(1/F) + (1/s)]^{-1} = [(1/{-100}) + (1/30)]^{-1 }= [(1/30) - (1/100)]^{-1}

s' = 42.86 mm. Since this is a positive number the image is real and to the right of the lens!

3. The total track, or the object to image distance, is 139 mm. What is the focal length required to have a 2x magnification? Where is the lens located relative to the object? (Magnification hint: the image distance is twice the object distance)

Solution: First let's consider the magnification... Technically I should have stated a -2x magnification.... We have two equations and two unknowns:

M = s'/s = -2 or s' = -2 * s (equation 1)

We're assuming the lens has not thickness. Total track or image to object is -s + s' =139 mm. The "-s" because it is a real object or to the left of the lens.

s = s' - 139 mm (equation 2)

Solve for s' by inserting equation 2 into equation 1

s' = -2 * (s' - 139) = -2s' + 278

3s' = 278

__s' = 92.7 mm__, positive and this is a real image to the right of the lens

Use equation 1 to solve for the object distance

s = - s' / 2

__s = -46.3 mm__, negative to the left of the lens and is a real object

Now solve for the focal length

1/s' = (1/F) + (1/s)

(1/F) = (1/s') - (1/s)

F = [(1/s') - (1/s)]^{-1} = [(1/92.7) - (1/{-46.3})]^{-1 }= [(1/92.7) + (1/46.3)]^{-1}

F = +30.9 mm. The lens is to the right of the object

4. (Extra Credit) Collimated light enters a ball of glass. What index of refraction will place the image on the back surface of the ball? (Hint: resembles a mirrors index)

Solution: one has to use the complete lens maker's equation that includes object and image space index of refraction for a single surface.

The back surface does not factor into the equation at all:

n'/s' = ((n'-n)/R) + (n/s)

Actually since the object is air we can use:

n'/s' = ((n'-1)/R) + (1/s)

With the object at infinity the term 1/s becomes zero, simplifying the equation

n'/s' = ((n'-1)/R)

Now it is algebraic manipulation to solve for n':

n'/s' = n'/R - 1/R

n'/s' - n'/R = -1/R or.... n'/R - n'/s' = 1/R

We want the image on the back surface so we know that s' = 2R

n'/R - n'/2/R = 1/R

n'/2/R = 1/R

n'/2 = 1

or n' = 2.0 : A spherical piece of glass with an index of 2.0 will image a distant object on its back surface.