We have lens formula, 1/f = 1/u + 1/v
where, f is focal length = +10 cm,
u is object distance = 5 cm and
v is image distance = ?
therefore using lens formula, 1/10 = 1/5 + 1/v ==> 1/v = 1/10 - 1/5 = -1/10
==> v = -10 cm [Answer] Negative image distance indicates image is virtual that is on the same side of the lens as the object.
and magnification = image distance / object distance
==>magnification = -10/10 = -1 [Answer] Negative magnification indicates image is inverted.
Consider a les that ha a focal hofAn object is locatem away fronm this lens. In...
An object is placed 12cm away from a convex lens that has a focal length of 5cm. a. Sketch a ray diagram for the situation. b. Is the image real or virtual, magnified or reduced, and upright or inverted? c. Determine where you should put a screen in order to project the image clearly. Is your answer consistent with the ray diagram? How do you know?
c. An object is placed 12cm away from a diverging lens with a focal length of -6am. i. (l pt) Use this information to caloulate the image distance and the magnification. i. (2 pts) Draw an accurate ray diagram for the above set-up using the rules for "easy" rays.
An object is 72 mm away from a converging lens with a focal length of 250mm. Another converging lens with a focal length of 200mm is placed 161mm to the right of the first. Draw the ray diagram to scale. Note: Magnification of final image should be 2.
A 10-cm tall object is placed 40 cm away from a lens that has a focal length of +8 cm. (a) What kind of lens is this? (b) Draw a ray diagram (c) Calculate image distance and height (I got a Di of 10 cm and a H of -2.5 but I can't match that with the diagram) (d) If you moved the lens a little to the right, the image would [ ] move to the right [ ]...
An object is located 23.0 cm to the left of a diverging lens having a focal length f = −37.2 cm. (a) Determine the distance and location of the image. (b) Determine the magnification of the image. (c) Construct a ray diagram for this arrangement.
Problem 5) a) An object is 80 cm to the left of a lens of focal length +40 cm. Where is the image? Is it real or virtual? Is it erect or inverted? What is its magnification? Draw the ray diagram. (2pt.) b) A second lens, of focal length +50 cm, is 120 cm to the right of the first lens. What is the distance between the final image and the first lens? Draw the ray diagram. (2pt.) 1
The focal length of a convex lens is given to be 17 cm. It's magnification is -2.5. (a) What is the distance of the object and image from the lens? (b) What is the image height if the object is 15 cm? (c) Draw a ray diagram for this situation, label completely.
A 20 cm tall object is located 70 cm away from a diverging lens
that has a focal length of 20 cm. Use a scaled ray tracing to
answer parts a-d.
a. Is the image real or virtual?
b. Is the image upright or inverted?
c. How far from the lens is the image?
d. What is the height of the image?
e. Now use the thin lens equation to calculate the image
distance and the magnification equation to determine...
A 0 cm tall object is placed 10 cm away from a concave mirror that has a 4.0 cm focal length. Calculate the: Image distance Image Height Magnification A 1.0 cm tall object is placed 5 cm away from a biconcave lens that has a 10.0 cm focal length. Calculate the: Image distance Image Height Magnification Using the optics as configured in #5 & #6, Draw a ray-tracing diagram, with all principle rays. What is the nature of the image?
A concave lens refracts parallel rays in such a way that they are
bent away from the axis of the lens. For this reason, a concave
lens is referred to as a diverging lens.
Part A
Consider the following diagrams, where F represents the focal point
of a concave lens. In these diagrams, the image formed by the lens
is obtained using the ray tracing technique. Which diagrams are
accurate?(Figure 1) (Figure 2) (Figure 3) (Figure 4)
Type A if...