A diverging lens with a focal length of -48.0 cm. forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation

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.

Consider a 2.5-cm tall candle positioned 7.5 cm in front of a concave spherical mirror with a radius of 30 cm. Determine and explain the characteristics of the resulting image. Please create a scaled ray diagram to aid in your explanation. Also, calculate the longitudinal magnification of the candle's image.

Thin lenses. Object O stands on the central axis of a thin symmetric lens. For this situation, each problem in the table (below) gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of the lens as object O or on the opposite side. (a) (b) (c) (d) (e) Lens i m R/V I/NI Side +14 C, 24 (a) Number Units (b) Number i Units (c) (d) (e) > >

A blue whale eyeball may be taken as a sphere that is about 15 cm in diameter. Assuming that it is filled with material with an index of refraction of 1.5, where would the image form for an object (in water) that is very far away? Just consider the initial image formed by refraction through the front surface. Select answer from the options below 15 cm behind the back of the eye 7.5 cm behind the front surface of the eye 15 cm behind the front surface of the eye Approximately 30 cm behind the back of the eye Approximately 50 cm behind the back of the eye

Please don't provide handwritten solution....   A diverging lens with f = -37.5 cm is placed 15.0 cm behind a converging lens with f = 21.0 cm .Where will an object at infinity be focused? Determine the image distance from the second lens. Follow the sign conventions.

An object is located 50.0 m to the left a converging lens of focal length 15.0 cm. A diverging lens of focal length 4.20 cm is placed 10.0 cm to the right of the converging lens. Locate (final image position) and describe (real or virtual, upright or inverted, reduced or enlarged) the final image of the object formed by the two-lens system.

Draw a principal axis through a convex lens. Label both items clearly. Label 2 focal points located 2.50 cm on each side of the lens. Draw and label a 1.0 cm-tall upright arrow sitting on the principal axis as object 6.00 cm from the lens. Construct two rays from the object, refracting correctly through the center of the lens. Draw and label the image formed.

A 1.0-cm-tall object is 70 cm in front of a converging lens that has a 30 cm focal length. Calculate the image position. Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.