An image is formed using a convex lens, the image being 15 cm past the lens. A second lens is placed 25 cm past the first lens and another image is formed, this time 10 cm past the second lens. Which of the following statements is true? The last image is inverted with regard to the original object. The last image must be larger than the object. The first image is virtual None of the above statements is true.

Using a single optic with a positive focal length, an image is created. As a result, which of the following MUST be true? Choose all that apply. The image is real The image is virtual The image is inverted O The image is upright The image distance is positive The image distance is negative The image height is positive The image height is negative The image has a positive magnification The image has a negative magnification The optic is converging The optic is diverging

For a thin lens situaiton, you have detemined the image distance is negative. Which of the following must be true?     the image is real, it is on the opposite side as the object     the image is virtual, it is on the same side as the object     the image is real, it is on the same side as the object     the image is virtual, it is on the opposite side as the object

An object is placed 40 cm from a converging lens with a focal length of 20 cm. The image formed by this lens becomes the object for a second diverging lens with a focal length of −15 cm. The distance between the two lenses is 50 cm. Calculate the final image distance formed by this two-lens system. Draw a diagram of the optical system.

You have a thin, diverging lens. If the value of q (the distance from the image to the mirror along the principal axis of the mirror) is -3.57cm and the distance of p (the distance from the object to the mirror along the principal axis of the mirror) is 17.16cm, what is the focal length of the diverging lens? Object Ray 1 Principal focal point Virtual image Diverging lens Ray 1 Ray 3 Ray 2 Secondary focal point |fl ·|fl· Note: Do not explicitly include units in your answer (it is understood the unit is cm). Enter only a number. If you do enter a unit, your answer will be counted wrong.

If a object is placed between a convex lens and its focal point, the image formed is O virtual and inverted. O virtual and upright. real and inverted. real and upright.

You use a lens to form an image. The object is 50 cm from the lens and the image is formed at a position of -14 cm. What is the focal length of your lens? f = cm What is the magnification of the lens? m = Is the image formed real or virtual? --- Is the image upright or inverted? Are you using a converging lens or a diverging lens?

A concave lens has a focal length of -7 cm. A candle with a height of 5 cm is 25 cm away from the lens. Find the location of the image. If the image is real give a positive answer and a virtual image should be a negative answer.

An object is 60.0 cm from a converging lens and the object is 1.00 cm tall. What is the poisition and height of the image if the focal length of the lens is 25.0 cm? The object is at 0.0233 cm and the height of the image will be 0.715 cm upright. The object is at 0.0233 cm and the height of the image will be 0.000388 cm inverted. The object is at 42.9 cm and the height of the image will be 0.000388 cm upright. The object is at 42.9 cm and the height of the image will be 0.715 cm inverted.

Q: You hold up a piece of paper by a lens and see the image of a candle projected on the paper. Is this image real or virtual? Q: Will the projected image show the candle right-side-up or upside-down? Q: You look through a concave (diverging) lens at a candle. Will the image you see be smaller or larger in size than the actual candle?