Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of refraction of n=1.50 and a glass plate with an index of n = 1.80, the diameter of the third bright ring is 0.740 mm. ▾ Part A If water (n=1.33) now fills the space between the lens and the glass plate, what is the new diameter of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light Express your answer with the appropriate units. ^ · C ?

Light enters a substance from air at 45.0° to the normal. It continues within the substance at 25.0° to the normal. What is the critical angle for this substance when it is surrounded by air? 25.2° O Othere is no critical angle O O 60.3° 90.0° 36.8° 45.2°

Part A A converging lens with a focal length of 60 cm and a diverging lens with a focal length of -50 cm are 290 cm apart. A 3.5-cm-tall object is 80 cm in front of the converging lens. Calculate the image position relative to the diverging lens. Express your answer to two significant figures and include the appropriate units. Enter a positive value if the image is on the other side from the lens and a negative value if the image is on the same side. HẢ ? s'2 Value Units %D Submit Request Answer Part B Calculate the image height. Express your answer to two significant figures and include the appropriate units. HA h' = Value Units %3D Submit Request Answer

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f1 = 39.4 cm. Precisely 36.3 cm behind this lens is a concave lens with a focal length f2 = -10.2 cm. The object to be photographed is 4.03 m in front of the objective lens. ▼ Q Search Part A How far behind the concave lens should the film be placed? Express your answer using two significant figures. 15| ΑΣΦ d = Submit Part B Request Answer What is the linear magnification of this lens combination? Express your answer using two significant figures. 15. ΑΣΦ P Pearson ? ? cm Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions Contact Us |

Luis is nearsighted. To correct his vision, he wears a diverging eyeglass lens with a focal length of -0.50 m. When wearing glasses, Luis looks not at an object but at the virtual image of the object because that is the point from which diverging rays enter his eye. Suppose Luis, while wearing his glasses, looks at a vertical 14-cm-tall pencil that is 2.0 m in front of his glasses.

Part A A stamp collector uses a converging lens with a focal length 50.0 mm to view a stamp 70.0 mm in front of the lens. Where is the image located? 175.0 mm You are correct. Your receipt no. is 168-5755 ? Previous Tries Part B Is the image real or virtual? Upright or inverted? Larger or smaller than the object? In front or behind the lens? V Real (R) or Virtual(V) I✓ Upright(U) or Inverted(I) L✓ Larger (L) or Smaller(S) F✓ In Front (F) or Behind (B) the lens An image is in front of the lens if it is on the same side as the object. Submit Answer Incorrect. Tries 3/6 Previous Tries

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