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What's the main problem with a pinhole lens?
It doesn't let much light through, so the image it forms is dim.
What does a lens do?
It changes the vergence of the light passing through it. Sometimes this forms an image.
What is an image?
An image is a 2D surface where each point on the surface says how much light is coming from a certain direction.
If an object moves closer to a lens, what happens to the divergence of the light entering the lens (i.e. what happens to \( V_{in} \))?
The divergence increases. (By increasing divergence, we mean becoming more negative, i.e. going from \( -2\text{D} \) to \( -4\text{D} \) )
If an object moves further from a lens, what happens to the divergence of the light entering the lens (i.e. what happens to \( V_{in} \))?
The divergence decreases. (By decreasing divergence, we mean becoming less negative, i.e. going from \( -4\text{D} \) to \( -2\text{D} \) )
If an object moves further from a lens, what happens to the convergence of the light leaving the lens ( \( V_{out} \))?
The convergence of the light leaving the lens will increase, because the divergence entering the lens decreases.
If the convergence of light leaving a lens increases, what happens to the position of the image?
The distance from the lens to the image will decrease, and the image will move closer to the lens. The image distance is \( 1/V_{out} \), and if \( V_{out} \) gets bigger, \( 1/V_{out} \) will get smaller.
If an object moves away from a lens, does the image move closer to or further from the lens?
It moves closer to the lens. When the image moves further away, the divergence \( V_{in} \) gets less, and so the convergence \( V_{out} \) of light leaving the lens increases. Thus the image gets closer.
What is the power \( F \) of a lens?
The power is the amount that the lens changes the vergence of light passing through it.
Is it possible for divergent light to leave a converging lens?
Yes it is possible.
If the divergence of the light entering the lens \( V_{in} \) exceeds the lens power \( F \), then the lens will not be able to converge it fully. For example, an object is placed \( 0.2\text{m} \) from a lens with power \( F=3\text{D} \). In this case, \( V_{in}=1/(-0.2)=-5\text{D} \), and so \( V_{in}+F = -5+3=-2\text{D} \). Thus the light leaving the lens has a vergence of \( -2\text{D} \), so it's still diverging, although not as much as the light entering the lens.
What is the definition of the image focal point?
The image focal point is the point where the image forms when parallel light enters the lens (i.e. when \( V_{in}=0 \)).
What is the definition of the object focal point?
The object focal point is where you have to place an object so that parallel light leaves the lens (i.e. to make \( V_{out}=0 \)).
When does divergent light leave a converging lens?
When the object distance is less than the focal length of the lens. Then \( V_{in} \) is more divergent than the power \( F \) can converge completely.
How can you place an object to ensure that converging light will leave a converging lens?
You must put the object further away than the focal length of the lens.
What is the shortest possible distance from lens to image?
It is the image focal length.
The further away an object is placed, the closer the image gets to the lens. If you place an object infinitely far away, so that \( V_{in}=0 \), then the thin lens equation says that \( V_{out}=V_{in}+F=F \). The image distance is then \( 1/V_{out}=1/F=f \).
An object is placed \( 0.4\text{m} \) to the left of a \( +4\text{D} \) lens. Where does the image form?
The vergence entering the lens \( V_{in} \) is \( 1/(-0.4)=-2.5\text{D} \). From the thin lens equation,
The image distance is then \( 1/V_{out}=1/1.5 = 0.667\text{m} \) (rounded to 3 decimal places). This is to the right of the lens.
An object is placed \( 80\text{cm} \) to the left of a \( +5\text{D} \) lens. Where does the image form?
All distances must be in metres, so \( 80\text{cm} \) is \( 0.8\text{m} \). The vergence entering the lens \( V_{in} \) is \( 1/(-0.8)=-1.25\text{D} \). From the thin lens equation,
The image distance is then \( 1/V_{out}=1/3.75 = 0.267\text{m} \) (rounded to 3 decimal places). This is to the right of the lens.
An object is placed \( 90\text{cm} \) to the right of a \( +3\text{D} \) lens. Where does the image form?
In this case, the light is travelling from right (the object) to left (the lens), so rightwardsis the positive direction. This means that the distance from lens to the image on the right is negative.
All distances must be in metres, so \( 90\text{cm} \) is \( 0.9\text{m} \). The vergence entering the lens \( V_{in} \) is \( 1/(-0.9)=-1.111...\text{D} \). From the thin lens equation,
The image distance is then \( 1/V_{out}=1/1.888... = 0.529\text{m} \) (rounded to 3 decimal places). This is to the left of the lens.
An image forms \( 0.75\text{m} \) to the right of a \( +2\text{D} \) lens. Where is the object?
The vergence leaving the lens \( V_{out} \) is \( 1/0.75=+1.333...\text{D} \). The thin lens equation says
Filling in what we know, we have \( V_{in}+2=1.333\text{D} \)
, so
\( V_{in}=1.333...-2 = -0.666...\text{D} \)
. The object distance is then
\( 1/V_{in}=1/-0.666... = -1.5\text{m}
\) . This is to the left of the lens.
An image forms \( 0.5\text{m} \) to the right of a \( +6\text{D} \) lens. Where is the object?
The vergence leaving the lens \( V_{out} \) is \( 1/0.5=+2\text{D} \). From the thin lens equation,
Filling in what we know, we have \( V_{in}+6=2\text{D} \) , so \( V_{in}=2-6 = -4\text{D} \). The object distance is then \( 1/V_{in}=1/-4 = -0.25\text{m} \) . This is to the left of the lens.
An object is placed \( 450\text{mm} \) to the left of a lens, and an image forms \( 20\text{cm} \) to the right of the lens. What is the power of the lens?
The object distance is \( -0.45\text{m} \) and the image distance is \( 0.2\text{m} \). The thin lens equation says
Filling in what we know, \( 1/(-0.45)+F = 1/0.2 \), so
\( F=1/0.2 - 1/(-0.45) = +7.22\text{D} \).
An object is placed \( 4000\text{mm} \) to the left of a lens, and an image forms 0.4m to the right of the lens. What is the power of the lens?
The object distance is \( -4\text{m} \) and the image distance is \( 0.4\text{m} \). The thin lens equation says
Filling in what we know, \( 1/(-4)+F = 1/0.4 \), so
\( F=1/0.4 - 1/(-4) = +2.75\text{D} \).