enter the answer in mm 11.957 not 1.1957.
part c. image is erect.
part d. object and image are on same side.
statement in the question right of the lens seems not correct as diverging lens forms image on the same side . we take -ve for image distance meaning on the left of the lens
A diverging lens with a focal length of -47.0 cm forms a virtual image 7.50 mm...
Item 6 Constants Part A A diverging lens with a focal length of -47.0 cm forms a virtual image 8.50 mm tall, 17.0 cm to the right of the lens. Determine the position of the object. cm Submit Reque st Answer Part B Determine the size of the object. Submit Request Ans
1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long...
A converging lens with a focal length of 12.2 cm forms a virtual image 8.10 mm tall, 18.0 cm to the right of the lens. Determine the position of the object. Determine the size of the object.
A converging lens with a focal length of 12.0cm forms a virtual image 8.00mm tall. Image is 17.0cm from the lens. Calculate (a)the position of the object, (b)magnification of the lens, (c)height of the object. (d)Is the image upright or inverted? (e) Are the object and image on the same side or opposite sides of the lens? (f) Draw ray-diagram
A converging lens with a focal length of 11.6 cm forms a virtual image 7.70 mm tall, 16.0 cm to the right of the lens. Part A Determine the position of the object. Express your answer in centimeters to three significant figures. VALO o ? S= 6.72 cm Part B Determine the size of the object. Express your answer in centimeters to two significant figures. J V AEO ? [y] = cm
A 4-cm tall object is placed 59.2 cm from a diverging lens having a focal length of -29.5 cm. a) Is the image produced by this lens virtual or real? b) Is the image inverted or upright? c) Is the image on the same side of the lens as the object or on the opposite side as the object? d) Where is the image located? (Please provide the magnitude of the position, no negative numbers) e) How tall is the...
HW Help A converging lens with a focal length of 11.6 cm forms a virtual image 7.70 mm tall, 16.0 cm to the right of the lens. f= 11.6 cm image height (y') = 7.70 mm image distance (s') = -16 cm object distance (s) = 6.72 cm A. Determine the size of the object. y? I keep getting the wrong answer. It's not 18.33 or 3.24. Please show work! Thank you!
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 converging lens with a focal length of 6.50 cm forms an image of a 3.80 mm -tall real object that is to the left of the lens. The image is 2.10 cm tall and erect. Where are the object and image located? in cm
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens. Part A What is the focal length of the lens? Part B Is the lens converging or diverging? Part C If the object is 8.00mm tall, how tall is the image? Part D Is it erect or inverted?