A simple refracting telescope with an objective lens with focal length 2.00m and and eye piece of with focal length of 2.0 cm is used to view Mars when it is 6 x 107km from the Earth. For the purposes of this question assume the unaided human eye can resolve objects 0.002 radians apart, what feature size can be seen on Mars with telescope?
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- Your camera’s zoom lens has an adjustable focal length ranging from 55 to 150 mm. Part (a) What is the maximum power of the lens, Pmax, in diopters? Part (b) What is the minimum power of the lens, Pmin, in diopters?Needs Complete typed solution with 100 % accuracy.The refracting telescope at the Yerkes Observatory has a 1.00-m diameter objective lens of focal length 20.0 m. Assume it is used with an eyepiece of focal length 2.50 cm. (a) Determine the magnification of Mars as seenthrough this telescope. (b) Are the Martian polar caps right side up or upside down?
- A large reflecting telescope has an objective mirror with a10.0 m radius of curvature. What angular magnificationdoes it produce when a 3.00 m focal length eyepiece isused?When the muscles connected to the crystalline lens relax, the focal length is 9.0000 cm. With this focal length, how close must an object be to form sharply focused images on the retina? (Note: this distance is called the near point of vision.)Needs Complete solution with 100 % accuracy don't use chat gpt or ai plz plz plz plz plz send.
- A compound microscope is made from an objective lens of focal length fo = 8 mm, andan eyepiece of fe = 12 mm. The objective lens is moved close to the object so that anintermediate image is formed 20 cm above the objective lens; then the eyepiece is set atits own focal length above the intermediate image, forming a virtual image at infinity forthe user’s eye. If the object is 0.15 mm in diameter, calculate the following, taking care todefine symbols used:i) The required distance of the objective lens above the object.ii) The diameter of the intermediate image.iii) The angular size of the image to the user’s eye.iv) The system angular magnification, defined as the angular size in (iii) divided by theangle subtended by the object to the naked eye from a distance of 25 cm.MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Your physics study partner is nearsighted. She has a near point of 50.0 cm from her eyes and she wears eyeglasses designed to allow her to read a newspaper held at a distance of 28.5 cm from her eyes. Determine the focal length of her glasses for each of the following cases. (a) She wears her eyeglasses 1.80 cm from her eyes. cm (b) She wears her eyeglasses 2.80 cm from her eyes. cmUse the worked example above to help you solve this problem. The near point of a patients eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 28.5 cm away? Neglect the eye-lens distance. 50 x cm (b) What is the power of this lens? 2 diopters (c) Repeat part (b), taking into account the fact that, for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye. 2.26 X diopters
- A small telescope has a concave mirror with a 2.6 m radius of curvature for its objective. Its eyepiece is a 3.4 cm focal length lens. Part (b) What angle (in degrees) is subtended by a 25,000 km diameter sunspot? Assume the sun is 1.50 × 108 km away. Part (c) What is the image angular size (in degrees) in this telescope?6. A 2.0 cm tall object is 15 cm in front of a converging lens that has a 20 cm focal length. (a) Use ray tracing to find the position and height of the image. To do this accu- rately, use a ruler or a paper with a grid. Determine the image distance and image height by making measurements on your diagram. (b) Calculate the image position and height. Compare with your ray-tracing an- swers in part a.