A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.
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- A lens in air has a convex front face with a radius of curvature of 12cm and a concave back face with a radius of curvature of 28cm. The index of refraction of the lens is 1.60. A 5cm tall object is placed 45cm in front (to the left) of the lens. A second lens with the same focal length is placed 315cm to the right of the first lens. What is the size of the final image? (Give the height in centimeters with the correct sign)In a darkened room, a burning candle is placed 1.63 m from white wall. A lens is placed between candle and wall at a location that causes a larger, inverted image to form on the wall. When the lens is moved 86.6 cm toward the wall, another image of the candle is formed. K 1.63 m How far apart are the two object distances that produce these images? Answer in units of m. Find the focal length of the lens. Answer in units of m.A contact lens is made of plastic with an index of refraction of 1.40. The lens has an outer radius of curvature of +1.90 cm and an inner radius of curvature of +2.42 cm. What is the focal length of the lens? Answer in cm
- A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 4.50 m to the right of the slide. The image is 60.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principalray diagram.On one side of a converging lens of focal length 27.6 cm, you position an object of height 3.09 cm somewhere along the principal axis. The resultant image has a height of 14.21 cm. How far from the lens is the object located?
- When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.20 cm-tall object is 60.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300 cm to the right of the first lens along the same optic axis. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The eye's lens. Part A Find the location and height of the image (call it I₁) formed by the lens with a focal length of 40.0 cm. Enter your answers in centimeters separated by a comma. s₁, y = 120,2.40 cm Submit Correct Set Up: Apply Solve: Lens 1: m₁ = Previous Answers Part B $1 f₁ $1-f₁ $₂,|y₂| = 1 S s' 1 S +1/1 = = gives (60.0 cm) (40.0 cm) 60.0 cm-40.0 cm to each lens. m₁ 120 cm 60.0 cm = = -2.00 = Y'₁ Y1 120 cm and m2 $1…A transparent photographic slide is placed in front of a converging lens with a focal length of 2.29 cm. An image of the slide is formed 11.6 cm from the slide. (a) How far is the lens from the slide if the image is real?A convex lens has a focal length of 13 cm. A candle with a height of 7 cm is 7 cm away from the lens. Find the location of the image. If the image is real give a positive answer and a virtual image should be a negative answer.
- An object is placed 30.5 cmcm from a +5.0-DD lens. A spherical mirror with focal length 25 cmcm is placed 75 cmcm behind the lens. Note that the mirror reflects light back through the lens. Find the distance from original object to the final image, measured from the object toward the lens. Enter negative value if the object is between the image and the mirror and positive value otherwise.A jellyfish is floating in a water-filled aquarium 1.46 m behind a flat pane of glass 3.90 cm thick and having an index of refraction of 1.5. (The index of refraction of water is 1.33.) (a) Where is the image of the jellyfish located? (Find the location of the final image, taking the inside and outside surfaces of the pane of glass as two refracting surfaces. Use a negative number to indicate the image is in the tank. Measure the distance from the air- glass interface. Enter your answer to three decimal places.) (b) Repeat the problem when the glass is so thin that its thickness can be neglected. (Use a negative number to indicate the image is in the tank. Enter your answer to three decimal places.)You are designing a lens to be made of glass with index of refraction 1.70. The first surface (the surface toward the object) is to be convex with radius of curvature 28.0 cm, and the focal length of the lens is to be 14.0 cm. (a) What must be the radius of curvature of the second surface (the surface away from the object)? (b) Will the second surface be concave or convex?