ImageObject1 y6.00*LOpticaxisLensPart DIs it real or virtual?virtualrealSubmitRequest Answer (Figure 1) shows an object and its image formed by a thinlens. Assume that L=16.2 cm and y = 3.45 mm.FigureImageObject6.00cmL1 of 1OpticaxisLensPart AWhat is the focal length of the lens?Express your answer in centimeters.- ΑΣΦf =SubmitPart BWhat type of lens is it?SubmitRequest AnswerconvergingdivergingPart CIIRequest AnswerWhat is the height of the image?Express your answer in millimeters.Π ΑΣΦ=??cmmm

We use the fomrula, to find the focal length of the lens. Given that Hence, the focal length is…

Answered: (Figure 1) shows an object and its…

Similar questions

Part A A certain lens focuses light from an object 1.55 m away as an image 41.3 cm on the other side of the lens. What type of lens is it? Charing O converging O diverging gs Is Submit Request Answer Part B What is its focal length? Follow the sign conventions. Express your answer to three significant figures and include the appropriate units. HA f- Value Units Submit Request Answer Part C MacBook Air 00 F3 000 40) F4 F5 F6 F7 FB F9 F10 F11 F12 & 5 6 7 8 9 E T D F G P R %24 キ3
Needs Complete typed solution with 100 % accuracy. Don't use chat gpt or ai i definitely upvote you.
An object with a height of 2.36 cm is placed 35.4 mm to the left of a lens with a focal length of 34.0 mm. Part A Where is the image located? ► View Available Hint(s) di = ΠΫΠΙ ΑΣΦ VO Submit Previous Answers ? m
Solve it correctly please. I ll ra
A glass sphere (n = 1.50) with a radius of 16.0 cm has a tiny air bubble 4.30 cm above its center. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere? 7.936 Apply the equation relating the object distance, image distance, and radius of curvature for a spherical refracting surface. cm Need Help? Read It Master It
How do I go about the diagram drawing? The lamp is placed 6.7 cm from convex lens 1 that has a focal distance of 2.5 cm. This lens is placed2.5 cm from another lens that also has a focal distance of 2.5 cm. Locate the final image of the object and theover-all magnification by completing the chart below. Show all work.Draw a ray diagram to verify the image’s placement.
look at picture
IP An object is located to the left of a concave lens whose focal length is -41 cm . The magnification produced by the lens is mị = 0.32. Part A To decrease the magnification to m2 = 0.23, should the object be moved closer to the lens or farther away? closer farther away Submit Request Answer Part B Calculate the distance through which the object should be moved. Express your answer using two significant figures. ΑΣφ Ad, = cm Submit Request Answer Provide Feedback |0
- Part A Consider a typical convex passenger-side mirror with a focal length of -80 cm. A 1.7-m-tall cyclist on a bicycle is 25 m from the mirror. You are 1.4 m from the mirror, and suppose, for simplicity, that the mirror, you, and the cyclist all lie along a line. How far are you from the image of the cyclist? Express your answer with the appropriate units. μΑ L = Value Units Submit Request Answer Part B What is the image height? Express your answer with the appropriate units. HA h' = Value Units %3D
True statements for optics Which of the following statements for single optic devices are true? Choose all that apply. It turns out that convex lenses always have a positive focal length. For real images, the image distance is negative and the object distance is positive. By convention, an inverted image has a positive image height. A magnification of -2 means the image is magnified. 0 Experiments have shown that converging optics can create virtual images. If an image has a magnification of -8, then the image is virtual. Which must be true? Using a single optic, an image with a negative image height 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 The image is upright The image distance is positive The image distance is negative The image has a positive magnification The image has a negative magnification The optic is converging The optic is diverging The optic has a positive focal length…
Part A How far from a 51.0 mm focal-length lens must an object be placed if its image is to be magnified 2.60× and be real? do = = Submit Part B do = ΑΣΦ What if the image is to be virtual and magnified 2.60×? Submit Request Answer VE ΑΣΦ Request Answer ? ? mm mm
IF YOU CAN NOT ANSWER BOTH DO NOT ANSWER ALSO ANSWER IN 3 SIG FIG USING SCIENTIFIC NOTATION USING E

SEE MORE QUESTIONS