A plano-convex lens (flat on one side, convex on the other) with index of refraction n rests with its curved side (radius of curvature R) on a flat glass surface of the same index of refraction with a film of index nfilm between them. The lens is illuminated from above by light of wavelength λ. Show that the dark Newton rings that appear have radii of
where m is an integer.
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Chapter 24 Solutions
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- A plano-convex lens has index of refraction n. The curved side of the lens has radius of curvature R and rests on a flat glass surface of the same index of refraction, with a film of index nflim between them, as shown in Figure P36.42. The lens is illuminated from above by light of wavelength . Show that the dark Newtons rings have radii given approximately by r=mRnfilm where r R and m is an integer. Figure P36.42arrow_forwardLight enters a prism of crown glass and refracts at an angle of 5.00 with respect to the normal at the interface. The crown glass has a mean index of refraction of 1.51. It is combined with one flint glass prism (n = 1.65) to produce no net deviation. a. Find the apex angle of the flint glass. b. Assume the index of refraction for violet light (v = 430 nm) is nv = 1.528 and the index of refraction for red light (r = 768 nm) is nr = 1.511 for crown glass. For flint glass using the same wavelengths, nv = 1.665 and nr = 1.645. Find the net dispersion.arrow_forwardBy ray tracing or by calculation, find the place inside the glass where rays from S converge as a result of refraction through the lens and the convex air-glass interface. Use a ruler to estimate the radius of curvature.arrow_forward
- What is Brewster’s angle for light traveling in water that is reflected from crown glass?arrow_forwardIf a microscope can accept light from objects at angles as large as =70 , what is the smallest structure that can be resolved when illuminated with light of wavelength 500 nm and (a) the specimen is in air? (b) When the specimen is immersed in oil, with index of refraction of 1.52?arrow_forwardAt what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank?arrow_forward
- Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared with the incident ray, what happens to the refracted ray? (a) It bends toward the normal. (b) It is undeflected. (c) It bends away from the normal.arrow_forwardTwo rays travelling parallel to the principal axis strike a large plano-convex lens having a refractive index of 1.60 (Fig. P23.54). If the convex face is spherical, a ray near the edge does not pass through the local point (spherical aberration occurs). Assume this face has a radius of curvature of R = 20.0 cm and the two rays are at distances h1 = 0.500 cm and h2 = 12.0 cm from the principal axis. Find the difference x in the position where each crosses the principal axis. Figure P23.54arrow_forwardFigure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin 1 = n2 sin 2. For paraxial rays, we assume 1 and 2 are small, so we may write n1 tan 1 n2 tan 2. The magnification is defined as M = h/h. Prove that the magnification is given by M = n1q/n2p. Figure P23.28arrow_forward
- Exp1ain why an object in water always appears to be at a depth shallower than it actually is?arrow_forwardHow many times will the incident beam in Figure P34.33 (page 922) be reflected by each of the parallel mirrors? Figure P34.33arrow_forward(4. Find the critical angle for lotal Internal on the surface Reflection of the light incident between the air and the diamond at point P a.) sindc = sin-i иг ni (sind) = 1.00 2.419) sint 0₁ = 24.4⁰° nz n₁ point = 1.333 2.419 n, sind, = n₂ sin da 35° 0 b.) If the diamond is totally immersed IN H₂0, find the Critical angle at the Diamond - Water interface. sin' (sinds)= ( sin 33° (2.419) sin(+6)= (1.33) sin O₂ sint (sin od= air 12.419 (sin (1160) 1.33 (1²60)) s.nl airnz 1.000 Diamond (n) = 2.6419 water = 1.333 ·Diamond A Oc= 33.4° C.) assuming the light ray remains perpendiular on entering the diamond, at what angle would the Light begin to emerge at point & if the diamond is rotated about point (Ⓒ of retraction) use Snell! diamond in the water (33.39⁰) 25°-33.4: 1.6° Todran > 33.4 Ог if & were a bit less 1 than 33.4 Like 33.3 -187.07° 2 the the Or in water w would be 85.07 or even 33.39 would give Or of 87.07 + 0,= $ 12,9° / + 2arrow_forward
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