What is the focal length of a drop (ball) of benzene (n = 1.501) of radius 2 mm. Determine the properties of the image resulting from an object 0.5 mm high and 5.3 cm from the center of the drop.
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What is the focal length of a drop (ball) of benzene (n = 1.501) of radius 2 mm. Determine the properties of the image resulting from an object 0.5 mm high and 5.3 cm from the center of the drop.
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- A diverging lens has a focal length of -26.7 cm. The image formed has a magnification of 0.62. What is the object distance? Express this numeric value in cm.If the distance of the object is 250 cm and the distance of the image is 140 cm, I say that the focal length is 390 cm. true or falsed = 2 / 4 | f = -4cm 2. The figure shows a diverging mirror with a focal length of -4cm. An object is 4cm to the left of the mirror and is 2cm tall. object b. Find the magnification of the image. c. Find the height h, of the image. makada wa k ho h₂ = a. a. Fill in the blanks above, and find d, the image-to-lens distance. समयमा 67% + @ matatatata ਲਈ ਲੋਕ ਅ daripada diverging mirror SEMUA ALLERGO heen Ladk STREET PERSE ASTERSUN dime APRESEN Ename E Para 6 d. Draw the focal point(s) of this mirror on the picture above. e. Draw the image above at the correct location and with the correct size & orientation.
- An object (height = 7.7 cm) and its image are on opposite sides of a converging lens. The object is located 14.0 cm from the lens. The image is located 5.6 cm from the lens. Determine the image height (in cm). Enter the numerical part of your answer to two significant figures. Hint: Remember that the sign of the image height is significant.- 25 cm. An object of height 2.8 cm is placed at 28 cm in front of a diverging lens of focal length, ƒ = Behind the diverging lens, there is a converging lens of focal length, f = 25 cm. The distance between the lenses is 3 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is cm. (Use the sign to indicate which side the image is on; positive sign di1 = means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is cm. (Use the sign to indicate which side the image is on.) d₁2 = c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is cm. (Use the sign to indicate whether the image is upright (positive) or hil = inverted (negative).) = d. How…Spherical refracting surfaces. When an object is placed 8.8 cm in front of a spherical refracting surface the image distance is -13 cm. The index of refraction of the refracting material is 2.7 and it is embedded in transparent material with index of refraction 1.8. Find (a) the radius of curvaturer of the surface (including the sign) and determine whether the image is (b) real or virtual and (c) on the same side of the surface as object O or on the opposite side. 2 n₁ n2 p 1.8 2.7 +8.8 (a) -13 (b) (c) R/V Side
- Can you please explain how...well...how this works. Can you draw a diagram or something? It seems like if you use m=v/u and enter 25 cm for v, then you're saying that the image is 25cm from whatever the rest of the system is in the equation. m is the magnification of the lens, so when you arrive at the value for u, that seems like it should be saying that if you put the iron 8.3 cm from the lens, then it will project the image 25 cm from the lens. Not 25 cm in front of your eyes. I don't understand how the answer ends up applying to the distance to the eyes. Is there some algebra that needs to be done to link the distance from the image to the distance to the eyes? By the way, what do v and u stand for? I know they're distances, but what words do the letters represent?Spherical refracting surfaces. When an object is placed 9.3 cm in front of a spherical refracting surface the image distance is -15 cm. The index of refraction of the refracting material is 2.7 and it is embedded in transparent material with index of refraction 2.0. Find (a) the radius of curvature r of the surface (including the sign) and determine whether the image is (b) real or virtual and (c) on the same side of the surface as object O or on the opposite side. (a) (b) (c) n1 n2 pr i R/V Side 2.0 2.7 +9.3 -15 (a) Number Units (b) (c) >A lab technician with a near point of 27 cm is viewing a specimen with a compound microscope that has a barrel length of 16.6 cm and an eyepiece that has a focal length of 1.5 cm. Determine the focal length the objective must have so that the overall magnification of the microscope will be -291. cm
- An object is placed 40 cm in front of a thin lens. If an image forms at a distance 120 cm from the lens, on the same side as the object, what is the focal length of the lens, in cm? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement.The figure shows an object and its image formed by a thin lens. Assume that d1 = 0.781 m, d2 = 0.604 m, and h = 7.50E-2 m. What is the image height (in m; answer sign and magnitude)?Two thin lenses with focal lengths of magnitude 15.0 cm, the first diverging and the second converging, are placed 12.00 cm apart. An object 3.00 mm tall is placed 5.50 cm to the left of the first (diverging) lens. Where is the image formed by the first lens located? Please provide a detailed explanation of the process. How far from the object is the final image formed? Please describe the steps taken to reach to your conclusion.