Draw a ray diagram for the diverging lens with the object height of 4 cm at the distance of 8 cm away. The focal point is at 3 cm. Find the image.
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- Draw a ray diagram for the diverging lens with the object height of 4 cm at the distance of 8 cm away. The focal point is at 3 cm. Find the image.
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- Where should an object be placed in front of a convex lens to get a real image of the size of the object?(a) At the principal focus of the lens(b) At twice the focal length(c) At infinity(d) Between the optical centre of the lens and its principal focus.An object of height 3 cm is placed at a distance of 25 cm in front of a converging lens of focal length 20 cm, to be referred to as the first lens. Behind the lens there is another converging lens of focal length 20 cm placed 10 cm from the first lens. Find the location, orientation, and size of the final image.An object is 0.02 m tall is placed 0.06 m in front of a diverging lens having a focal length of 0.04 m. Find (a) the image distance (b) the image size
- an object of height 5 cm is placed 20 cm in front of a converging lens at focal length 10 cm. Behind the converging lens, and 25cm from it, there is a diverging lens of the focal length of 6 cm. Find the location of the final image, in centimeters, with respect to teh diverging lens. what is the magnification of the final image? what is the height of the final image?Find the position of the image of a small object given by a glass sphere (n=1.4) of radius 4 cm, knowing that the object is located in the air (n=1) at 21 cm from the center of the sphere.A spherical concave lens has a focal length of 50 cm, and an object is placed 25 cm from the lens. Draw a digram. Estimate the image distance Is it real or virtual upright or inverted magnified or reduced
- 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.A 3 cm tall object is placed 16 cm from a converging lens with a focal length of 12 cm. A diverging lens with a focal length of 10 cm is placed 36 cm behind the converging lens. Both lenses have the same principal axis. Draw the ray diagram to graphically find the final image. Need only handwritten solution only (not typed one).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
- 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.A concave lens has a focal length of -7 cm. A candle with a height of 9 cm is 24 cm away from the lens. Find the magnification of the image. If the image is upright give a positive answer and an inverted image should be a negative answer.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.