The linear magnification of an object placed on the principal axis of a convex lens of focal length 30 cm is found to be + 2. In order to obtain a magnification of -2, by how much distance should the object be moved?
Q: A concave lens forms a virtual image 0.5 times the size of the object. The distance between object…
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Q: A converging lens has a focal length of 10 cm. An object is placed 12 cm from the lens. (a) Using…
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A: Given that the value of p and q then we have to determine the value of focal length of the lens in…
Q: An object is placed 17 cm in front of a converging lens with focal length 10 cm. A second lens,…
A: Given The first lens having the focal length is f1 = 10 cm. The second lens having the focal length…
Q: A converging lens with a focal length of 40 cm and a diverging lens with a focal length of -40 cm…
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Q: Two convergent lenses, L1 (focal length = 15.6 cm) and L2 (focal length = 10.2 cm) are placed a…
A: 2 convex lenses distance "L" apart, Focal lengths, f1 = 15.6 cm f2 = 10.2 cm Object distance, u1 =…
Q: An object is placed 12.1 cm to the left of a diverging lens of focal length -6.97 cm. A converging…
A: Given: The diverging lens focal length is f1=-6.97 cm, The distance of the object to the left of a…
Q: A converging lens with a focal length of 15 cm is used to form an image of an object placed 25 cm…
A: Focal length of the coverging lens (f) = 15 cmObject distance (u) = 25 cm
Q: A compound microscope is made with an objective lens (fo = 0.90 cm) and an eyepiece (fo = 1.1 cm).…
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Q: An object is placed 12.0 cm to the left of a lens. divergent with a focal length of 6.00 cm. a…
A: For diverging lens: Object distance, u = -12.0 cm Focal length, f = -6.00 cm (sign convention)…
Q: A concave lens of focal length 9.7 cm has an object located on its principal axis, 41.8 cm from the…
A: Given, lens type is concave Focal length, f= 9.7 cm Object distance , u=41.8 cm Image distance v=?…
Q: An object with a height of -0.060 m points below the principal axis (it is inverted) and is 0.180 m…
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Q: An object has a height of 0.067 m and is held 0.220 m in front of a converging lens with a focal…
A: Heights of object =0.067mObject distance =0.22mFocal length =0.19m
Q: The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective…
A: The total angular magnification for a compound microscope is given as m=-L-fef0×Nfe Here f0 and fe…
Q: A 5.37 mm high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal…
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Q: An object has a height of 0.051 m and is held 0.220 m in front of a converging lens with a focal…
A: Given that, Height of the object, ho=0.051 m Object distance, u=0.220 m The focal length of the…
Q: A 1.00-cm-high object is placed 4.40 cm to the left of a converging lens of focal length 8.05 cm. A…
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Q: You are designing a symmetric double convex lens that is made of glass with index of refraction…
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Q: A 7.25 mm high pushpin sits upright in front of a converging lens at a distance of 25.4 cm. If the…
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Q: An object with a height of -0.060 m points below the principal axis (it is inverted) and is 0.110 m…
A: Height of object h1 = -0.060 mDistance of object u = 0.110 mFocal length of lens f = -0.35 mLet…
Q: A diverging lens has a focal length of magnitude 17.8 cm. (a) For an object distance of 44.5 cm,…
A: The focal length of the diverging lens is given as, f = 17.8 cm (i) The first object distance is…
Q: An object of height 7.25 cm is placed at a distance of 22.5 cm from the convex lens. Whose focal…
A: Given Type of lens is convex the focal length of lens is 10.5 centimetre the height of object is…
Q: A candle is placed at a fixed distance D from a screen. A converging lens can be put between the…
A: Given thatD = 87 cmdistance of candle from lens is x
Q: An object is located 15.4 cm to the left of a diverging lens having a focal length f = −35.8 cm.…
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Q: An object is located 15.8 cm to the left of a diverging lens having a focal length f = −36.2 cm.…
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Q: An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.200 m…
A: Height of object h1 = -0.040 m Distance of object u = -0.200 m Focal length of lens f = -0.27 m Let…
Q: An object is placed in 5 cm in front of a lens, producing an image that is upright and 3 times…
A: Given, m = +3 u = - 5 cm
Q: An object is placed at a distance of 10.0 cm from a convex lens with a focal length of 15.0 cm, and…
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- An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.140 m in front of a diverging lens. The focal length of the lens is -0.25 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? mA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 cm. A diverging lens of focal length –16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position 7.5 cm in front of the second lens v 0.5466 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtualConsider a converging lens with focal length 7.05 cm. The distance between an object and a real image of the object created by the lens is 50.5 cm. Find the distance between the object and the lens if the lens is closer to the object than it is to the image. Answer in cm.
- The distance between the eyepiece and the objective lens in a certain compound microscope is 10.4 cm. The focal length of the objective is 0.430 cm, and that of the eyepiece is 1.40 cm. Find the overall magnification of the microscope. (The near point of the eye is 25 cm. Assume that the object is placed at the focal point of the objective lens, and one places the eyepiece at the near point of the eye.)An object of 1 cm tall is placed 3cm in front of a converging lens of focal length of 2 cm. (a) Use ray tracing to find the image. (b) Use the lens equation to find the image distance and compare it to your ray tracing and find the percentage error.An object is 6 cm in front of a converging lens with a focal length of 10cm. Draw a ray diagram (to scale with a ruler) to find the location of the image. Is the image upright or inverted, and Is the image real or virtual? Then I want to use the thin lens formula to find the image distance and the magnification. I got stuck in the middle of this problem and am confused. Thank you for the help!
- The projection lens in a certain slide projector is a single thin lens. A slide 23.0 mm high is to be projected so that its image fills a screen 1.90 m high. The slide-to-screen distance is 3.06 m. (Enter your answers to at least one decimal place.) (a) Determine the focal length of the projection lens. mm(b) How far from the slide should the lens of the projector be placed to form the image on the screen? mmYou are tasked with sourcing a lens which will be made out of glass with refractive index 1.668, and is submersed in a liquid with refractive index 1.379. You want a focal length of 6.5 cm, and the radius of curvature of the first side of the lens is 7.4 cm. What radius of curvature is required for the second side, in cm? [Note the first/second lens surface radius is conventionally positive/negative for convex.]An object is placed 21.2 cm in front of a diverging lens which has a focal length with a magnitude of 13.0 cm. How far in front of the lens should the object be placed in order to produce an image that is reduced by a factor of 3.15?