EDT 4 - Optics and Optical Instruments. Solutions

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Dwain Stokes
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1 EDT 4 - Optics and Optical Instruments Solutions. μ sin30 0 = μ 2sin r (Snell s law at st interface) μ 2sin r = μ 3sin45 0 (Snell s law at 2 nd interface) 30 0 μ Hence, μ sin30 0 = μ 3sin45 0 μ sin30 0 = μ 3sin45 0 μ 3 = μ sin30 0 sin45 0 μ 3 = 2. When image is at near point, r r 45 0 μ 2 μ 3 m = ( + D f e ) L f o Putting the values we get L = 20cm. 3. In case of option 2, if size of object is large compared to wavelength of light, all the wavelengths are scattered nearly equally. Other options are obvious. 4. As image is formed on screen, it is real image and hence inverted. Also image is four times the object. Hence, m = -4 u = -x & v = 00 x m = v u Hence, 00 x x x = 20cm. = 4 u = -20cm & v = 80cm f = v - u Hence f = 6cm. 5. Same as that of Q4. x 00 cm 6. For end A: u = -0cm & f = 20cm v = + f u Hence v = -20cm. For end B: u = -20cm & f = 20cm v = + f u Hence v = -60cm. B B 5 cm 0 cm A 60 cm 20 cm A

2 Hence image of end A say A is formed at distance 20cm from lens while that of B say B is formed at distance 60cm from lens. Therefore length of image = 60cm 20cm = 40cm 7. For first case, object distance is u and image distance is v. u + v = f.. () For second case, object distance is 3u and image distance is 5v. 3u + 5v = f From () and (2), u + = v 3u + 5v Hence, 2 3u = - 4 5v = - 0 v 2v.. (2) So f = u + v = u - 0 2u = 6u Hence f = 6u. As u is negative, focal length is negative and hence mirror is Concave. 8. For lens: u = -30cm v = + f u = + 20 ( 30) 30 cm 35 cm v = 60cm. So image formed by the lens is at 60cm from lens on right side of it. This image is beyond the mirror. So for mirror, it will act as a virtual object. Object distance in that case will be positive and will be (60-35)cm i.e. 25cm. For mirror: u = +25cm v = - f u = - ( 00) 25 v = -20cm. Hence final image will be formed at distance of 20cm from mirror towards left i.e. at distance of 5cm from lens and in between mirror and lens. 9. Informative. 0. u + v = f Differentiating both the sides, u 2 du v 2dv = 0

3 Hence, dv = v2 u 2 du = m2 du This formula can be used to find size of image along principle axis (dv) if length of object and image both are very small compared to u and v. Here du is the length of object which is kept along principle axis. Using mirror formula, we can get that v = -30cm. Hence m = v u = 2 Hence dv = ( 2 )2 du = 0.5mm. Real image is formed on the other side of element e. Means light rays are passing through e to meet each other on other side. So e must be a lens. Also magnification m = -2 & u = -60cm. v u = -2 Hence v = (-2)(-60) = 20cm. Using lens formula, = f v - = - u 20 ( 60) Hence f = 40cm. 2. When object is a virtual object, i.e. when converging rays of light are incident on optical element, object distance is taken as positive as it appears that object is kept on the other side. 3. Component of velocity parallel to mirror will not contribute anything to the velocity of object. Component of velocity perpendicular to the mirror = Vsinθ. So image will move towards mirror with Vsinθ and w.r.t. ground as mirror itself is moving, velocity of image will be 2Vsinθ. 4. Assume object is at distance x from principle axis at some time. As it is at center of curvature, image will also be formed at center of curvature. Hence magnification is - and so image will also be formed at distance x from principle axis as Image Height = Object Height. So principle axis moves away from object at speed u and image moves away from principle axis at same speed i.e. u. So velocity of image is 2u w.r.t. ground. 5. Apparent distance of object from mirror = (00-30) v = f u = ( 30) ( 90) Hence v = -45cm. So image is formed at distance of 45cm from the mirror. = 90cm. x x Object Image u

4 6. Refractive index of material is 2. Hence Critical angle θ C = sin - (/ 2) = As final light ray is grazing the surface, it must be incident at critical angle. Hence r = 45 0 r = A r = 30 0 μ = sini sinr sin i = 2 x (/2) = / 2 i = Consider particle to be at distance r from one of the face. Then from other face it is at (5-r) Apparent Depth = Real Depth/μ. Hence r/μ = 7 & (5-r)/μ = 5. Solving these equations we get, μ = 5/2 = Informative. 9. For red color: δ r = (μ r - )A For violet color: δ v = (μ v - )A Hence, angular separation between two color δ = δ v δ r = (μ v μ r)a = Informative. 2. Incoming ray is just totally internally reflected. So it must be incident at critical angle. sin i = μ 2/μ Also for internal reflection at second interface, sin(90 i) > μ 3/μ cos i > μ 3/μ cos 2 i > (μ 3/μ ) 2 sin 2 i > (μ 3/μ ) 2 (μ 2/μ ) 2 > (μ 3/μ ) μ 2 < μ μ Informative. Medium i i 90 - i 90 - i Medium 3 Medium Let the two angle of incidences be i and i 2. As for both of them deviation is same, when angle of incidence is i, angle of emergence will be i 2 and vice-versa. This is because light rays can be traced backward. δ = i + e A 45 = i + i 2 60 i + i 2 = 05 i - i 2 = 5. Given i = 60 0 & i 2 = 45 0

5 24. Refer question number Use the concept of refraction at spherical surfaces. μ = 3/2, μ 2 =, u = -5cm, R = -0cm μ 2 v - μ = μ 2 μ u R Putting the values, we get v = -4cm. So apparent depth is 4cm. 26. Person is suffering from Short-sightedness. Corrective lens should be such that if an object is at infinity, its image should be formed at 200cm. So focal length must be -200cm as lens has to be concave. Finally, power will be given by /f = 0.5D 27. There will be no refraction at first boundary as it is normal incidence. So angle of incidence at second boundary will be Also, as sin 45 0 = / 2 = /.4. So green and blue light are incident with an angle greater than critical angle while red is incident at an angle less than critical angle. Hence, green and blue light will undergo total internal reflection and come out from the bottom side while red will escape from the hypotenuse. 28. As images formed in two cases are of same size and one of them is real while other is virtual, magnifications in two cases can be taken as m and m. Hence m = v u = - v 2 u 2 From lens formula, f = v u = u ( u v ) So, u f = () m Similarly, Adding () and (2), f = (u + u 2 ) 2 u 2 f = (2) m 29. Let height of bird as perceived near the mirror when viewed from water be h B. Then h B = yμ + h Depth of image in the mirror = d I = h B = yμ + h Depth of image as seen by the bird d IB = yμ+h + y = 2y + h μ μ Velocity of image w.r.t. bird = d dt dib = d dt (2y + h μ ) = 2 dy dt = 2V 30. μ = 3/2, μ 2 = 4/3, R = 0cm μ 2 v - μ = μ 2 μ u R Putting the values we get, v = 2 3 ( u ) Bird y h

6 Image is real only if it is formed in region Y i.e. if v > 0. So we get that u > 0 => u < -80cm. So image will be virtual if distance between object and surface is less than 80cm. 3. Assumes powers of two lenses are 3P and -5P respectively. Effective power = 3P 5P = /0.6-2P = 0.6 P =.2 Hence individual powers are: 3P = 0.4 and -5P = So focal lengths are: 40cm and -24cm. 32. Informative. 33. Parallel beam incident on convex lens will form image at its focus. If this image is formed at the focus of concave lens, rays will emerge parallel to each other. So distance between lenses should be sum of the focal lengths of two lenses. d = 30cm + (-0cm) = 20cm. 34. m = f v f = f f u. For the first case, magnification is m when object distance is u while for the second case, magnification is m when image distance is 2u. So we get: f m = f u = f 2u f. Dividing numerator and denominator by f, m = r = 2r where r = u f Solving for r, we get r = 3/2 or r = 0. The case r = 0 will correspond to a spherical mirror where f i.e. when it is almost a plane mirror. Using values of r, we get values of m as -2 or. 35. As angle of incident is same as that of emergence, this is the case of minimum deviation. δ = δ min = i + e A = = 30 0 μ = sin(a+ δ min ) 2 sin( A 2 ) = 36. Informative. 37. Informative. 38. Informative. 39. Use lens formula. sin (60) sin (45) = All the colors having refractive index more than that of yellow, will get totally internally reflected. So only Orange and Red will make it into the air.

7 4. From lens maker s formula, f = (μ - )( ) = ( 3 ) ( ) = R R Hence f = 20cm. 42. Informative. 43. Graph of angle of deviation v/s angle of incidence is not symmetric about point of minimum deviation. So angle of minimum deviation will be closer to 37 0 than to So most possible value is δ 44. Informative. 45. Use mirror formula and formula for magnification. 37 δ min 53 i

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UNIT-5 EM WAVES UNIT-6 RAY OPTICS

UNIT-5 EM WAVES UNIT-6 RAY OPTICS UNIT-5 EM WAVES 2 Marks Question 1. To which regions of electromagnetic spectrum do the following wavelengths belong: (a) 250 nm (b) 1500 nm 2. State any one property which is common to all electromagnetic