When do (object distance) is very large, what does the thin lens equation predict for the value of 1/f?
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1. An object is placed a distance s to the left of a thin convex lens with focal length f. Use the thin lens equation, 1 /s + 1 /s' = 1/ f , to determine for which ranges of s the image is (a) Real and magnified (larger than the object) (b) Real and reduced (smaller than the object) (c) Virtual
An object is located at a distance of 6 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 4 cm from the converging lens and 10 cm from the object. The diverging lens has a focal length of -3 cm. Note: To handle a multiple lens system, we treat them independently. We first find the image created by the first lens. We then use the image from the first lens to act...
An object is on the left side of a thin converging lens. The object is located at a distance of 6 cm away from a thin converging lens with focal length of 2 cm. Use the thin lens equation (1/f = 1/s' + 1/s) to predict the following: (a) Location of the image? (b) Magnification of the image (including inverted versus non-inverted)? (c) Real or virtual? Draw diagram please!
(a) Find the object distance (in terms of f) for a thin converging lens of focal length f if the image is real and the image distance is seven times the focal length. (Use any variable or symbol stated above as necessary. Enter your coefficient to two decimal places.) p = (b) Find the object distance (in terms of f) for a thin converging lens of focal length f if the image is virtual and the absolute value of the...
1. Using the thin lens equation (1/f=(1/s)+(1/s')), prove that the focal length of a lens is equal to the image distance when there is parallel incident light 2. There is a specific case where for a single converging lens the object distance equals the image distance. Find the relation between the focal length and the object distance in this case.
(2) An object is located at a distance of 4 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 3 cm from the converging lens and 7 cm from the object. The diverging lens has a focal length of -2 cm. Use the thin lens equation to predict the following (a) Location of the final image? to the object? (c) Is the final image real or virtual?
Use the thin lens equation to prove that an object at infinity produces an image at the focal length of the lens (i.e. that when p=infinity, q=f) Note that q=12.5 cm
A thin lens is imaging an object. The total distance between the object and its image is 1m. If the lens has a focal length of 0.2m, find the location of the object from the lens.
You are provided a convex (+) thin lens and an object where a real image at a focal length f=f is found. Find the minimum distance possible from the object to the image using the focal length. Draw a ray sketch and use the thins lens equation.
An object is imaged on to a photographic sensor using a thin positive lens. The distance between the object and its image is 18 cm, such that the image size is half the size of the object. (a) Determine the distance between the lens and the image on the photographic sensor. (b) A glass plate having plane parallel faces, is 0.5 cm thick and has a refractive index of 1.5. This is inserted between the lens and the photographic sensor,...