Lecture 32 - Lenses and Optical Instruments

Thin lenses

A lens makes use of the refractive properties of a material to focus light.

The basic building blocks of lenses are faces that are either spherical or planar.

The spherical faces can either curve outwards (convex) or inwards (concave).

The key property of a lens is that it focuses rays traveling parallel to the axis to a particular point, the focal point.

The focal point is a distance $f$ from the center of the lens and is defined as the focal length of the lens.

A related measurement of the focusing ability of a lens is the power $P$

$P=\frac{1}{f}$

measured in Diopters ($\mathrm{D}$) which are equivalent to $\mathrm{m^{-1}}$

Converging lens

A lens which is thicker in the center than at the edges is a converging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=f$ from the center of the lens.

To obtain a parallel beam of light light should be radially propagating outwards from the point $F'$ a distance $x=-f$ from the center of the lens.

Converging Lens applet.

Diverging Lens

A lens which is thinner in the center than at the edges is a diverging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=-f$ from the center of the lens, which means that the rays appear to diverge outwards from that point.

To obtain a parallel beam of light light the incoming rays should have a path such that they would go to the point $F'$ a distance $x=f$ from the center of the lens if the lens were not there..

Diverging Lens applet.

Raytracing for a converging lens

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

1. A ray that leaves the object parallel to the axis and then goes through $F$
2. A ray that passes through the center of the lens and is not bent
3. A ray that passes through $F'$ and exits parallel to the axis

Raytracing for a diverging lens

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

1. A ray that leaves the object parallel to the axis and then goes through $F$
2. A ray that passes through the center of the lens and is not bent
3. A ray that passes through $F'$ and exits parallel to the axis

Lens equation

Ray tracing can be a pain, so it is useful to have a lens equation like we had for mirrors that links the important quantities, $d_{o}$, $d_{i}$ and $f$.

From the orange similar triangles

$\frac{h_{i}}{h_{o}}=\frac{d_{i}-f}{f}$

and from the green ones

$\frac{h_{i}}{h_{o}}=\frac{d_{i}}{d_{o}}$

so

$\frac{d_{i}-f}{f}=\frac{d_{i}}{d_{o}}\to\frac{1}{f}-\frac{1}{d_{i}}=\frac{1}{d_{o}}$

which leads to the lens equation

$\frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}$

This equation also works for diverging lens (with $d_{i}$ and $f$ negative).

As with mirrors, the magnification $m$ is

$m=\frac{h_{i}}{h_{0}}=-\frac{d_{i}}{d_{o}}$

Sign Conventions:

• The focal length is positive for converging lenses and negative for diverging lenses.
• The object distance is positive if the object is on the side of the lens from which the light is coming (this should mostly be the case, but there are exceptions when there are combinations of lenses).
• The image distance is positive if the image is on the opposite side of the lens from which the light is coming (real image) and negative otherwise (virtual image).
• The height of the image is positive if upright and negative if inverted with respect to the object.

Two converging lenses

Combinations of lens can be treated sequentially, by first finding the image produced by the first lens and then then using it as the object for the next lens.

Applying the lens equation to the first lens

$\frac{1}{d_{iA}}=\frac{1}{f_{A}}-\frac{1}{d_{oA}}$

and to the second lens

$\frac{1}{d_{iB}}=\frac{1}{f_{B}}-\frac{1}{d_{oB}}$

and then using $d_{oB}=l-d_{iA}$ allows the determination of the final image position.

The first lens produces an image that has height $-\frac{d_{iA}}{d_{oA}}h_{o}$, which will then be used as the object height in the next lens, so the final image has height

$\frac{d_{iA}}{d_{oA}}\frac{d_{iB}}{d_{oB}}h_{o}$

When considering the multiplying power of lens combinations we can simply multiply the effects of the individual lens.

Lensmaker equation I

How can we get from our laws of refraction to a rule for how to make a lens with a particular focal length?

The answer is through the lensmaker equation.

Here we will consider in the simplified case of a thin lens.

Throughout our derivation we will frequently use approximations that are only valid when the thickness of the lens is significantly less than the radius of curvature.

The first step is to apply Snell's law to the two interfaces

$\theta_{1}\approx n \theta_{2}$
$\theta_{4}\approx n \theta_{3}$

Then we find 3 triangles which give us

$\theta_{1}\approx\sin\theta_{1}\approx\frac{h_{1}}{R_{1}}$     $\alpha\approx\sin\alpha\approx\frac{h_{2}}{R_{2}}$     $\beta\approx\tan\beta\approx\frac{h_{2}}{f}$

We can also find from inspection that

$\gamma=\theta_{1}-\theta_{2}$      $\theta_{4}=\alpha+\beta$

and from the angles around the point at the tip of the $h_{2}$ line

$(\frac{\pi}{2}-\alpha)+\theta_{3}+(\frac{\pi}{2}-\gamma)=\pi\to\alpha=\theta_{3}-\gamma$

Our goal is to now combine these equations to arrive at a an equation in terms of the lens radii and refractive index which is what a lensmaker can control.

Lensmaker equation II

1. $\theta_{1}\approx n \theta_{2}$
2. $\theta_{4}\approx n \theta_{3}$
3. $\theta_{1}\approx\sin\theta_{1}\approx\frac{h_{1}}{R_{1}}$
4. $\alpha\approx\sin\alpha\approx\frac{h_{2}}{R_{2}}$
5. $\beta\approx\sin\beta\approx\frac{h_{2}}{f}$
6. $\gamma=\theta_{1}-\theta_{2}$
7. $\theta_{4}=\alpha+\beta$
8. $\alpha=\theta_{3}-\gamma$

Using 8, and subbing in 6, and 2

$\alpha=\frac{\theta_{4}}{n}-(\theta_{1}-\theta_{2})$

Now using 7

$\alpha=\frac{\alpha}{n}+\frac{\beta}{n}-\theta_{1}+\theta_{2}$

and then 1

$\alpha=\frac{\alpha}{n}+\frac{\beta}{n}-\theta_{1}+\frac{\theta_{1}}{n}$

We now use equations 3,4 and 5 to write

$\frac{h_{2}}{R_{2}}=\frac{h_{2}}{nR_{2}}+ \frac{h_{2}}{nf}-\frac{h_{1}}{R_{1}}+\frac{h_{1}}{n_{R_{1}}}$

We once again turn to the thin lens approximation and say $h_{1}\approx h_{2}=h$ which allows us to cancel out $h$, and on multiplying both sides by $n$ and rearranging we get our equation

$\frac{1}{f}=(n-1)(\frac{1}{R_{1}}+\frac{1}{R_{2}})$

The eye

In order to understand optical instruments we need to consider how our eyes process incoming light from objects to visualize them.

The first refraction for incoming light occurs when light passes from air to the cornea (n~1.376), and the lens (n=1.386-1.406) is adjusted by the muscles around it to accommodate objects at different distances.

The eye has a near point which is the shortest distance at which it can focus on an object. This varies from person to person, and with age, but is typically on the order of 25cm.

Corrective Lenses

Nearsightedness or myopia is an inability to focus on distant objects, the eye focuses the light to a point in front of the retina.

If the rays are diverged by a diverging lens before they reach the eye then the rays are able to be focused on the retina.

Farsightedness or hyperopia is an inability to focus on near objects, the eye focuses the light to a point behind the retina.

If the rays are converged by a converging lens before the enter the eye then the rays can be focused on the retina.

Perceived size of an object

The apparent size of an object, and the amount of detail we can resolve, depends on the angle $\theta$ the object subtends on our retina.