Version 087 EX4 ditmire (58335) 1

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Bruno Lewis
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1 Version 087 EX4 ditmire (58335) This print-out should have 3 questions. Multiple-choice questions ma continue on the next column or page find all choices before answering. 00 (part of ) 0.0 points A material with an index of refraction of.34 is used to coat glass. The index of refraction of glass is.5. What is the minimum thickness of the coating that will minimize the reflection of light with a wavelength of 4640 Å? Correct answer: µm. Phase Changes for Reflecting Waves. Since the coating has a refraction that is less than that for glass (but greater than that for air), we know that the reflected light from both the glass and the coating will undergo a 80 phase shift. This means that the total trip inside the coating must be exactl one half of the wavelength of light inside the coating. (Hence the waves reflected from the glass and from the coating will interfere destructivel.) The wavelength of light inside the coating is λ where n is the index of refraction of the coating. This implies that the n thickness of the coating must be one fourth the wavelength of light inside this medium. Hence (calling the thickness of the material t, the wavelength of light in air λ, and the index of refraction of the medium n) t λ 4 n m (4) (.34) t m µm. 00 (part of ) 0.0 points Now assume that the coating s index of refraction is.56. Assume that the rest of the sstem (from the previous question) remains the same. What is the minimum thickness of the coating needed to minimize the reflection of this light now? Correct answer: µm. This part is reall solved in the same wa as Part except now the index of refraction of the material is greater than that for glass. Hence the light reflected from the material surface undergoes a 80 phase transition, but the light reflected from the glass within the material goes through a 0 phase transition. This means that the light must travel a full wavelength within this material in order to interfere destructivel. Using the same notation as before, we then sa: t λ n m (.56) t m µm points Consider the setup of a single slit experiment. Hint: Use a small angle approximation; e.g., sin tan.

2 Version 087 EX4 ditmire (58335) a S 0 3 Determine the height 3, where the third minimum occurs λ a. 3 7 λ a λ a λ a 5. 3 λ a 6. 3 λ a λ a correct λ a λ a 0. 3 λ a The third minimum occurs at β 6 π, which corresponds to a path difference between two end ras: b 3 β k ( 6 π ) π λ 3 λ b 3 a 3 3 b 3 a 3 λ a points The filament of an incandescent lamp has a 50 Ω resistance and carries a direct current of.7 A. The filament is 3 cm long and 0.83 mm in radius. Calculate the Ponting vector at the surface of the filament Correct answer: W/m. et : I.7 A, l 3 cm 0.03 m, R 50 Ω, and r 0.83 mm m. The power P radiated b the filament is given b P I R. This power is distributed uniforml over the surface area of the filament (Area π r l), so the intensit S of the radiation at the surface equals S P A I R π r l (.7 A) (50 Ω) π ( m) (0.03 m) W/m.

3 Version 087 EX4 ditmire (58335) (part of ) 0.0 points Consider a light ra which enters from air to a liquid, where the index of refraction of the liquid is given b n and the index of refraction of air is given b n. n n light ra Air iquid Consider the following three ratios, where each is defined b the specified quantit in the liquid,,, and c, to that in the air, λ, f, and c. What is the ratio of their wavelengths, λ λ? λ 4 λ 4 λ λ λ λ λ correct λ 3 λ λ The frequenc of an electro-magnetic wave is independent of the media in which it is present; that is, f. A ra with a frequenc f has a wavelength λ c in the vacuum. In a medium with an f index of refraction n, and from the definition of the index of refraction, n c c. So for n c f ) ( c f ( c ) ( ) n f ( ) ( ) c n f λ n. λ n, f, and c c n. 006 (part of ) 0.0 points What is the ratio of their frequencies, f? f correct f 4 f f 4 f f f f f 3

4 Version 087 EX4 ditmire (58335) 4 0. f The frequenc of an electro-magnetic wave is independent of the media in which it is present; that is, f. Therefore, f points ight strikes the surface of a prism, n.84, as shown in the figure. 45 If the prism is surrounded b a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection within the prism? Correct answer: Basic Concept: Given: Solution: sin c n r n i n i.84 c 45 sin c n r n i n r n i (sin c ).84(sin 45 ) points A double-slit interference experiment is shown below. 0.5 mm S m If the third order bright fringe falls a distance of 5 mm from the center of the pattern, what is the wavelength of the light used? Correct answer: nm. d S Q et : m d 0.5 mm λ nm S Q 90 r r ( tan ) δ d sin r r P O

5 Version 087 EX4 ditmire (58335) 5 Third order, m 3 λ d m ml (0.5 mm )(5 mm ) 3 ( m ) nm points ight from a distant star is collected b a concave mirror. How far from the mirror is the image of the star if the radius of curvature is 55 cm? Correct answer: 77.5 cm. two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope,.4 m in diameter, is in orbit 97.7 km above Earth and is turned to look at Earth. If ou ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 473 nm? Correct answer:.349 cm. p + q f m h R h q p Concave Mirror f > 0 >p> f f <q < 0 >m> f >p> 0 <q < 0 >m> et : r 55 cm. Incident light from an infinitel distant object will strike the mirror as parallel ras, and will converge at the focal point, which is half the radius of curvature of the mirror. Hence d i f r 77.5 cm points The resolution of a lens can be estimated b treating the lens as a circular aperture. The resolution is the smallest distance between w For single slit diffraction x λ w, For a circular aperture the resolution differs b a factor of.. Therefore x. λ w. (97700 m) ( m) (.4 m) (00 cm/m).349 cm. This resolution is so poor that NASA cannot read license plate numbers on automobiles, but it has been reported that NASA is x

6 Version 087 EX4 ditmire (58335) 6 able to do this. Do ou believe this report, and if so, what technique is used? The answer to this question ma be classified points Consider the setup of a single slit experiment. a image that is 5.86 cm in height. a) What is the magnification of the image? S Correct answer: Determine the phase angle difference β between the two end ras, ra r and ra r, at the position of the first intensit minimum. et : h 3.0 cm h 5.86 cm. and. β 3 π M h h 5.86 cm 3.0 cm β 4 3 π 3. β π correct 4. β π 5. β 3 π 6. β 4 π 7. β π 8. β 5 π 9. β 5 π 0. β 7 π The first minimum is at β π, where β φ π, and φ π is the phase difference of the two ras for destructive interference. 0 (part of ) 0.0 points An object 3.0 cm tall is placed 3.9 cm in front of a mirror, which creates an upright 03 (part of ) 0.0 points b) What is the radius of curvature of the mirror? Correct answer: cm. R p + q et : p 3.9 cm. M q p q Mp, so

7 R p + q p + Mp ( R p + Mp [ (3.9 cm) ) + (.9404)(3.9 cm) cm. Version 087 EX4 ditmire (58335) 7 ] points Consider two lenses where f f and f f. As shown in the sketch, the are aligned along the z-axis with a separation d.5 f. 3f f # # The distance between the object and lens is p 3 f. An image is formed due to the two-lens sstem. Find the distance of this image from lens. d points What is the maximum radiation pressure that can be exerted b sunlight in space of intensit 3387 W/m on a flat black surface? The speed of light is m/s e e e e e e e e e e-05 Correct answer: Pa. et : I 3387 W/m c m/s. and The maximum pressure is attained when the light is normall incident on the surface. Under this circumstance, the relation between the intensit and the pressure on a black surface is given b P I c 3387 W/m m/s Pa.. q 3 f. q f 3. q 3 f correct 4. q 7 f 5. q f 6. q 4 f 7. q f 8. q 3 f 9. q 3 4 f 0. q 5 f For lens, the object is the image formed b lens located at q f to the right of lens. So in the formula f p + q, for the virtual object, p (q d) 4 f.

8 Version 087 EX4 ditmire (58335) 8 This leads to q f p f + 4 f f + 4 f Basic Concepts: 0 π/ π 3 π/ π φ π or f q f points A is illuminated b 58 nm light as shown in the figure below. The distance from the slits to the is 4.5 m. δ λ 58 nm φ π δ λ d sin d λ λ + d λ 0.73 mm S 44% 4.5 m Double slit interference the phase angle difference φ maxima: φ π, 4 π, 6, 8 π, minimum: φ π, 3 π, 5, 7 π. [ ] [ φ I E 4 E0 cos cos ω t + φ ] ( ) φ I I 0 cos, with φ k δ. () Figure: Not drawn to scale. Find the minimum positive phase angular value φ such that I I 0 44%, where I 0 is the intensit at the central maximum and I is the intensit at the position on the Correct answer: et : I I(φ) I 0 I(0) 44% 0.44 Solution: Using Eq., we have [ ] I φ arccos I 0 [ ] arccos 0.44 arccos( ) ( rad).6909 rad

9 Version 087 EX4 ditmire (58335) 9 Note: The difference in path length for the ras can be found as follows. Using δ λ φ, and Eq., we have π δ λ φ π [ I arccos λ π I 0 ] ( m) { [ ]} arccos 0.44 π { } (.6909) ( m) π ( m){ rad} m nm. Note: The angle can be found as follows: d S et : Q S Q 90 r r ( tan ) δ d sin r r P I 0.44 I 0 λ 58 nm m 4.5 m, and d 0.73 mm m. O { ( m) arcsin π ( m) [ ] } arccos 0.44 [ ( ] m) (.6909) arcsin π ( m) arcsin( ) rad Note: Since tan, and using Eqs. and, we have tan ( { [ ]}) λ I tan arcsin π d arccos I 0 ( { ( m) tan arcsin π ( m) [ ] }) arccos 0.44 ( [ ( ]) m) (.6909) tan arcsin π ( m) tan(arcsin[ ]) (4.5 m) tan( rad) (4.5 m) tan( ) m mm. 07 (part of ) 0.0 points Consider the setup of a single slit experiment. The wavelength of the incident light is λ 470 nm. The slit width and the distance between the slit and the is specified in the figure. Solution: Using Eq. and φ π δ λ d sin, we have { λ } λ φ arcsin π d { [ λ I arcsin π d arccos I 0 ]} () 540 µm S 8.9 m Find the position of the first intensit minimum. Use a small angle approximation sin tan.

10 Version 087 EX4 ditmire (58335) Correct answer: mm. et : λ 470 nm, 8.9 m, a 540 µm. and 540 µm S 8.9 m Find the intensit ratio I I mm a S [ δ a sin a ] For single slit diffraction, destructive interference occurs when, a sin λ, or simpl when, δ a sin λ. Thus, between the two end ras which correspond to the first minimum, the phase angle difference is β π and the path length difference is δ λ. The small angle approximation gives us tan sin δ a, or δ a λ mm. a 08 (part of ) 0.0 points Denote the intensit on the at mm b I and the intensit on the at 0 b I 0. Correct answer: et: I I 0 [ ] sin(β/). β/ 0 π π β δ π or 360 λ β π δ λ a sin λ R mm mm a λ () 0.43, ()

11 Version 087 EX4 ditmire (58335) where is the position of the first intensit minimum. From Eq. (), β π a [ λ ] π (540 µm) ( mm) 470 nm (8.9 m).7077 rad 54.8, in agreement with the above diagram Following is the alternative method. From Eq. () and (), we have R β β δ δ. β Rβ (0.43) π.7077 rad (since β π), the intensit ratio at an point on the is ) I I 0 sin ( β β [ ] sin(rπ) Rπ [ sin(0.43π) 0.43π 0.590, in agreement with the intensit diagram in the question points A narrow beam of light passes through a plate of glass with thickness.69 cm and a refractive index.5. The beam enters from air at an angle.5. The goal is to calculate the deviation d of the ra as indicated in the figure..5 ] Find deviation d Correct answer: cm. Note that s x cos s x, so x s cos Also: sin( ) d x Yielding d x sin( ) s sin( ) cos.69 cm sin ( ) cos cm 3 d n d.69 cm d n points A beam of white light is incident on a triangular glass prism with an index of refraction of about.5 for visible light, producing a spectrum. Initiall, the prism is in a glass aquarium filled with air, as shown.

12 Version 087 EX4 ditmire (58335) Aquarium Spectrum Aquarium Spectrum Incident ight Prism Red Violet Incident ight Prism Red Violet If the aquarium is filled with water with an index of refraction of.3, which of the following is true?. The position of red and violet are reversed in the spectrum.. The violet light disappear. Thus, the spectrum produced has less separation between red and violet than that produced in air points Two narrow parallel slits are illuminated with light of wavelength 460 nm. 3. The spectrum has the same separation between red and violet as that produced in air. 4. The spectrum produced has less separation between red and violet than that produced in air. correct 0.3 mm S 4 m 4 mm 5. The intensit of the light increases. 6. The spectrum produced has greater separation between red and violet than that produced in air. 7. There is no light come out of the prism. 8. No spectrum is produced. 9. The red light disappear. 0. A spectrum is produced, but the deviation of the beam is opposite to that in air. What is the phase difference between the two interfering waves on a at a point 4 mm from the central bright fringe? Correct answer: Since water has a larger index of refraction than air (but still smaller than the glass prism), the deviation of light beam caused b refraction will be smaller. et : 4 m 4 mm d 0.3 mm.

13 Version 087 EX4 ditmire (58335) 3 d 0 S Q S Q 90 r r ( tan ) δ d sin r r π or 360 φ P 0 π/ π 3 π/ π O in agreement with the above diagram. 0 (part of ) 0.0 points When Mars is nearest the Earth, the distance separating the two planets is km. Mars is viewed through a telescope whose mirror has a diameter of.6 cm. If the wavelength of the light is 73 nm, what is the angular resolution of the telescope?. 4.88e e e e e e e e e e-06 δ λ 460 nm φ π δ λ d sin d λ λ + d λ Solution: We obtain δ d sin ( ) d sin + ( ) (0.04 m) d sin (4 m) + (0.04 m) (0.3 mm) sin(0.00 rad) 30 nm, φ πd λ + π ( m) ( m) (0.04 m) (4 m) + (0.04 m) rad 4.609, Correct answer: rad. The Raleigh criterion gives for the angular resolution. λ D m 0.6 m rad. The smallest distance d that can be resolved between two points on Mars is d ( rad) ( m) km. 03 (part of ) 0.0 points What is the smallest distance that can be resolved between two points on Mars?

14 Version 087 EX4 ditmire (58335) Correct answer: km.

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : , 1 O P T I C S 1. Define resolving power of a telescope & microscope and give the expression for its resolving power. 2. Explain briefly the formation of mirage in deserts. 3. The radii of curvature of