An object is 13.8 cm from the surface of a concave (diverging) thin lens of focal...

Question

Question

science physics

Answer 1

Answer #1

- U = -13.8 cm - clemi t = ť t = 1 el at = of the ], S 13.8 v=-5.06 cm Asuer ma -S.O6 -13.8 0-37 Ansuer CS SCANNED withe Cam

Answer 2

Similar Homework Help Questions

A real object is 13.8 cm to the left of a thin, diverging lens having a...

A real object is 13.8 cm to the left of a thin, diverging lens having a focal length of magnitude 18.5 cm. (a) Is the sign of the focal length negative or positive? _________ (b) Find the image distance. _______cm (c) Find the magnification. _______
14. An object is 2 cm from a concave (diverging) lens. The resulting height of the...

14. An object is 2 cm from a concave (diverging) lens. The resulting height of the virtual image is half as large as the the height of the object. What is the focal length of the lens? (Hint: Use the information given to determine the lateral magnification, and from that determine the image distance. The use the thin lens equation.) (A) - cm (B) - cm (C) -1 cm (D) –2 cm
A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length...

A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm. Draw the ray diagram for this situation. What is the location and height of the image? Is the image real or virtual? • Draw an optical axis with the lens centered on the axis. Represent the object at the correct distance from the lens • Draw the three "special rays" from the top of the object • Extend the rays...

A real object is 13.6 cm to the left of a thin, diverging lens having a focal length of magnitude 24.5 cm.

A real object is 13.6 cm to the left of a thin, diverging lens having a focal length of magnitude 24.5 cm. (a) is the sign of the focal length negative or positive? negative positive (b) Find the image distance. (c) Find the magnification. (d) State whether the image is real or virtual. real virtual (e) State whether the image is upright or inverted. upright inverted
An object is located 23.0 cm to the left of a diverging lens having a focal...

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.
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a fo...

A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...

A negative or diverging thin lens has a focal length of 10 cm. a) An object...

A negative or diverging thin lens has a focal length of 10 cm. a) An object is placed 30 cm from the lens. Find the image distance and explain if the image is real or virtual. b) Explain if the image is upright or upside down and if it is smaller or larger than the object.
You have a concave (diverging) lens with a −30 cm focal length. The magnification produced by...

You have a concave (diverging) lens with a −30 cm focal length. The magnification produced by the lens for a particular object distance is m = 1/2 and you wish to decrease the magnification to m = 1/8. Determine the distance and direction (away from or closer to the lens) through which the object must be moved in order to accomplish this.
A 20 cm tall object is located 70 cm away from a diverging lens that has...

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...

An object is located at a distance of 6 cm from a thin converging lens with...

An object is located at a distance of 6 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 4 cm from the converging lens and 10 cm from the object. The diverging lens has a focal length of -3 cm. Note: To handle a multiple lens system, we treat them independently. We first find the image created by the first lens. We then use the image from the first lens to act...