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Two lenses made of kinds of glass having different indices of refraction n1 and n2 are cemented together to form an optical doublet. Optical doublets are often used to correct chromatic aberrations in optical devices. The first lens of a certain doublet has index of refraction n1, one flat side, and one concave side with a radius of curvature of magnitude R. The second lens has index of refraction n2 and two convex sides with radii of curvature also of magnitude R. Show that the doublet can be modeled as a single thin lens with a focal length described by
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Chapter 35 Solutions
Physics for Scientists and Engineers
- The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens for light incident from the left. (b) What If? After the lens is turned around to interchange the radii of curvature of the two faces, calculate the focal length of the lens for light incident from the left.arrow_forwardA converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answersarrow_forwardWhy is the following situation impossible? Consider the lensmirror combination shown in Figure P35.55. The lens has a focal length of fL = 0.200 m, and the mirror has a focal length of fM = 0.500 m. The lens and mirror are placed a distance d = 1.30 m apart, and an object is placed at p = 0.300 m from the lens. By moving a screen to various positions to the left of the lens, a student finds two different positions of the screen that produce a sharp image of the object. One of these positions corresponds to light leaving the object and traveling to the left through the lens. The other position corresponds to light traveling to the right from the object, reflecting from the mirror and then passing through the lens. Figure P35.55 Problem 55 and 57.arrow_forward
- A lamp of height S cm is placed 40 cm in front of a converging lens of focal length 20 cm. There is a plane mirror 15 cm behind the lens. Where would you find the image when you look in the mirror?arrow_forward(i) When an image of an object is formed by a plane mirror, which of the following statements is always true? More than one statement may be correct. (a) The image is virtual. (b) The image is real. (c) The image is upright. (d) The image is inverted. (e) None of those statements is always true. (ii) When the image of an object is formed by a concave mirror, which of the preceding statements are always true? (iii) When the image of an object is formed by a convex mirror, which of the preceding statements are always true?arrow_forwardThe end of a solid glass rod of refractive index 1.50 is polished to have the shape of a hemispherical surface of radius 1.0 cm. A small object is placed in air (refractive index 1.00) on the axis 5.0 cm to the left of the vertex. Determine the position of the image.arrow_forward
- Two lenses made of kinds of glass having different indices of refraction n1 and n2 are cemented together to form an optical doublet. Optical doublets are often used to correct chromatic aberrations in optical devices. The first lens of a certain doublet has index of refraction n1, one flat side, and one concave side with a radius of curvature of magnitude R. The second lens has index of refraction n2 and two convex sides with radii of curvature also of magnitude R. Show that the doublet can be modeled as a single thin lens with a focal length described by 1f=2n2n11Rarrow_forwardFigure P38.43 shows a concave meniscus lens. If |r1| = 8.50 cm and |r2| = 6.50 cm, find the focal length and determine whether the lens is converging or diverging. The lens is made of glass with index of refraction n = 1.55. CHECK and THINK: How do your answers change if the object is placed on the right side of the lens? FIGURE P38.43arrow_forwardIn Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forward
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