We will only discuss double concave diverging lenses as shown in
[link] . Concave lenses are thicker on the outside and thinner on the inside.
[link] shows a concave lens with light rays travelling through it. You can see that concave lenses have the opposite curvature to convex lenses. This causes light rays passing through a concave lens to
diverge or be spread out
away from the principal axis. For this reason, concave lenses are called
diverging lenses. Images formed by concave lenses are
always virtual.
Unlike converging lenses, the type of images created by a concave lens is not dependent on the position of the object. The image is
always upright, smaller than the object, and located closer to the lens than the object.
We examine the properties of the image by drawing ray diagrams. We can find the image by tracing the path of three light rays through the lens. Any two of these rays will show us the location of the image. You can use the third ray to check the location, but it is not necessary to show it on your diagram.
Drawing Ray Diagrams for Diverging Lenses
Draw the three rays starting at the top of the object.
Ray
travels parallel to the principal axis. The ray bends and lines up with a focal point. However, the concave lens is a
diverging lens, so the ray must line up with the focal point on the same side of the lens where light rays enter it. This means that we must project an imaginary line backwards through that focal point (
) (shown by the dashed line extending from
).
Ray
points towards the focal point
on the opposite side of the lens. When it hits the lens, it is bent parallel to the principal axis.
Ray
passes through the optical center of the lens. Like for the convex lens, this ray passes through with its direction unchanged.
We find the image by locating the point where the rays meet. Since the rays diverge, they will only meet if projected backward to a point on the same side of the lens as the object. This is why concave lenses
always have virtual images. (Since the light rays do not actually meet at the image, the image cannot be real.)
[link] shows an object placed at an arbitrary distance from the diverging lens.
We can locate the position of the image by drawing our three rays for a diverging lens.
[link] shows that the image of an object is upright. The image is called a
virtual image because it is on the same side of the lens as the object.
The image is smaller than the object and is closer to the lens than the object.
An object is placed 4 cm to the left of a diverging lens which has
a focal length of 6 cm.
What is the position of the image?
Is the image real or virtual?
Draw the lens, object, principal axis and focal points.
goes from the top of the object parallel to the principal axis.
To determine the angle it has when it leaves the lens on the other side, wedraw the dashed line from the focus
through the point where
hits the lens. (Remember: for a diverging lens, the light ray on the opposite
side of the lens to the object has to bend
away from the principal axis.)
goes from the top of the object in the direction of the other focal
point
. After it passes through the lens, it travels parallel to the
principal axis.
goes from the top of the lens, straight through the optical centre
with its direction unchanged.
Just like for converging lenses, the image is found at the position where
all the light rays intersect.
Draw the image at the point where all three rays intersect.
The distance to the object is 2,4 cm.
The image is on the same side of the lens as the object, and is upright. Therefore it is virtual. (
Remember: The image from a diverging lens is
always virtual.)
The properties of the images formed by converging and diverging lenses depend on the position of the object. The properties are summarised in the
[link] .
Image Properties
Lens type
Object Position
Position
Orientation
Size
Type
Converging
inverted
smaller
real
Converging
inverted
same size
real
Converging
inverted
larger
real
Converging
no image formed
Converging
upright
larger
virtual
Diverging
any position
upright
smaller
virtual
Summary of image properties for converging and diverging lenses
Diverging lenses
An object 3 cm high is at right angles to the principal axis of a concave lens of focal length 15 cm. If the distance from the object to the lens is 30 cm, find the distance of the image from the lens, and its height. Is it real or virtual?
The image formed by a concave lens of focal length 10 cm is 7,5 cm from the lens and is 1,5 cm high. Find the distance of the object from the lens, and its height.
An object 6 cm high is 10 cm from a concave lens. The image formed is 3 cm high. Find the focal length of the lens and the distance of the image from the lens.
Questions & Answers
A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?