Consider a thick equi-convex lens (made of a material of refractive index 1.5) of the type shown in Fig. 7.15 the magnitude of the radii of curvature of the two surfaces is 4.cm, the thickness of the lens is 1 cm and the lens is placed in air. Obtain the system matrix and determine the focal length and the positions of unit planes. 多 多 多 多 Fig. 7.15
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- Can you help me with the wrong. Please? Thank you.Two plane mirrors make an angle α = 54.0◦ between them. A ray of light incident on one of the mirrors is reflected and hits the second mirror. Find the angle β between the ray incident on the first mirror and the ray reflected off of the second mirror. (a) 36.0◦ (b) 72.0◦ (c) 54.0◦ (d) 24.0◦ (e) None of the above.One end of the strip of a plane mirror is fixed and the other end rests on the top of the small vertical rod. The length of the plane mirror strip is 25cm. A ray of light is incident on the mirror and reflected from the mirror and forms a spot on a screen 3m away from the mirror. Now, if the top of the rod is moved upwards 0.1 mm, then what will be movement of the spot? A 0.24 cm B 3.4 cm 5 cm 4.5 cm
- The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.00 mm, which also varies. We shall assume that the radil of curvature of its two surfaces have the same magnitude. (Note: The results obtained in the parts A, B and C are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The eye's lens. Part A Find the radii of curvature of this lens. Express your answer in millimeters. R= 7.0 mm Submit ✓ Correct Set Up: = (n − 1)(-). If R is the radius of the lens, then R₁ = R and R₂=-R₁ ² + ² = }· m = ² = -² Solve: = (n − 1) (₁ - ₁) = (n − 1) ( ²2 - ) = ²(n-¹) R=2(n-1)f=2(1.44) (8.00mm) = 7.04mm Part B If an object 12.0 cm tall is placed 26.0 cm from the eye lens, where would the lens focus it? Express your answer in…You want to make a symmetrical biconvex lens (same radii of curvature) to project an image of an object magnified five times on a wall 40 cm away from the lens. The material has an index of refraction equal to 1.60.(c) What is the radius of curvature of the lens faces? 12 cm 8 cm 4 cm 10 cm 2 cm 6 cm (a) What is the distance from the object to the wall? (b) What is the focal length of the lens? (a) 8 cm (b) 6,67 cm (a) 18 cm (b) 6,67 cm (a) 28 cm (b) 6,67 cm (a) 48 cm (b) 6,67 cm (a) 8 cm (b) 8,00 cm (a) 18 cm (b) 8,00 cm (a) 28 cm (b) 8,00 cm (a) 48 cm (b) 8,00 cm (a) 8 cm (b) 9,67 cm (a) 18 cm (b) 9,67 cm (a) 28 cm (b) 9,67 cm (a) 48 cm (b) 9,67 cmA lens is made of glass that has refractive index 1.60. In air, the lens has focal length +18.0 cm. What is the focal length of this lens if it is totally immersed in a liquid that has refractive index 1.42?
- A technician needs to verify that a parabolic mirror intended for use in a telescope has been properly manufactured. The mirror's focal length should be 226 centimeters. How high above the base of the mirror (in cm) should a point on the mirror's surface located 28 cm horizontally from the base be (ie, what should be the value of y at the location x = 28 cm)?You 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.]The lenses in the figures below are all made of glass of index 1.50 and have surfaces with radii of curvature of magnitude either 9.90 cm or 21.8 cm. Find the radii of curvature and the focal length of each lens. (a) (b) (c)