Practice: A biconvex lens has a focal length of 12 cm. If an object is placed 5 cm from the lens, where is the image formed? Is it real or virtual? Is it upright or inverted? What’s the height of the image if the object is 2 cm tall?

Concept #1: Thin Lens Equation

Practice: A biconvex lens has a focal length of 12 cm. If an object is placed 5 cm from the lens, where is the image formed? Is it real or virtual? Is it upright or inverted? What’s the height of the image if the object is 2 cm tall?

Concept #2: Lens Maker Equation

Example #1: What Types of Images Can be Formed by Lenses?

Example #2: Image Formation by a Biconcave Lens

The cornea of the human eye (the outtermost layer) acts like a thin lens, which is why imperfections in the cornea can lead to poor eyesight. The front and rear surfaces of the cornea is roughly spherical, with a radius of curvature of about 7.8 mm for the front and a radius of curvature of about 7.3 mm for the rear. What would the focal length of a typical cornea be?

An object is to the left of a thin lens. The lens forms an image on a screen that is 2.60 m to the right of the object. The height of the image is 2.50 times the height of the object. What is the focal length of the lens?

An object is placed 16 cm from a double concave thin lens (the index of refraction is 1.5) with the magnitude of the radius of curvature 20 cm on both sides. What is the magnification of the image?
a) –0.56
b) +0.89
c) –0.89
d) 0.56
e) –1.10

The cross-section of a glass lens with an index of refraction 1.5, is shown below. Determine the focal length f using the small angle approximation. Include both the magnitude and sign, which indicates whether this is a divergent or convergent lens.
1. 0.514286
2. 1.03196
3. -0.902762
4. -3.99724
5. -3.02298
6. 0.992568
7. -3.60345
8. 0.904751
9. 1.08561
10. -0.703276

A slide projector creates a 90.0 cm high image of a 2.0 cm tall slide. The screen is 300 cm from the lens. (a) How far is the slide from the lens? (b) What is the focal length of the lens?

In order to form a real image with a converging lens that is twice as large as the object, how far away from the lens should the object be placed? Assume the lens has focal length f.
(a) f
(b) 3f / 2
(c) 2f
(d) 2f / 3
(e) f / 2
(f) None of the above

An object that is 2 mm tall is placed 10 cm to the left of a thin lens that has f = +30 cm. What is the height of the image?

When a 4.0 mm tall object is 40.0 cm to the left of a lens, an upright image is formed that is 2.0 mm tall. Is the lens converging (f > 0) or diverging (f < 0)?
A. Converging
B. Diverging

Sally places an object 3 cm from a thin convex (converging) lens along its axis. The lens has a focal length of 4.5 cm. What are the respective values of the image distance and magnification?
a. -9 cm and +3
b. 18 cm and -3
c. +7.2 cm and -0.6
d. -18 cm and -3

Consider a divergent lens with a focal length f. An upright object is placed within the interval f and 2 f to the left of the lens. The corresponding image is
1. real, upright, and the same size.
2. real, upright, and reduced.
3. virtual, upright, and the same size.
4. virtual, upright, and reduced.
5. virtual, inverted, and reduced.
6. non-existant.
7. virtual, inverted, and enlarged.
8. virtual, upright, and enlarged.
9. virtual, inverted, and the same size.
10. real, upright, and enlarged.

A concave mirror with a radius of curvature of 1.4 m is illuminated by a candle located on the symmetry axis 3.4 m from the mirror. Where is the image of the candle?

A 4 cm tall object is placed 60 cm away from a converging lens of focal length 30 cm. What is the nature and location of the image? The image isa. Real, 2.5 cm tall, and 30 cm from the lens on the same side as the object.b. Virtual, 2.5 cm tall, and 30 cm from the lens on the side opposite the object.c. Virtual, 2.0 cm tall, and 15 cm from the lens on the side opposite the object.d. Virtual, 4.0 cm tall, and 60 cm from the lens on the same side as the object.e. real, 4.0 cm tall, and 60 cm from the lens on the side opposite the object.

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.What is the location x i of the image?a) x i = -3f / 5b) x i = 3f / 2c) x i = 3f / 2d) x i = f / 2e) x i = f / 5

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.Choose the correct description of the image in the situation above.a) The image is real and inverted.b) The image is virtual and inverted.c) The image is real and upright.d) The image is virtual and upright.e) There is no image formed.

An object (bold arrow) is located at a distance of 3 f / 2 (i.e. at x = 3 f / 2) in front of a diverging lens as shown in the figure. Note that f is positive, so the focal length of this lens is - f.
Suppose the object is now placed at x = - f. What is the magnification M?
a) M = 0
b) M = 1/2
c) M = 1

A convex lens forms a virtual image 4 times the size of the object. The object distance is 2.43333 cm. Find the focal length of the lens.1. 2.43752. 10.63. 2.406254. 2.156255. 2.666676. 7.07. 2.933338. 3.244449. 0.79210. 1.77778

The magnitudes of the radii of curvature for the spherical surfaces A and B are, respectively, |RA| = a and |R B| = 2a. The material of which the lens is made has an index of refraction n = 1.5. Find the focal length of the thin lens.1. f = a2. f = 2a3. f = −2a4. f = 4a5. f = −3a6. f = 3a7. f = −a8. f = −4a