NQA6B tủ tài liệu training

NUMERICALS-ANSWERS QUESTIONS The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm What is the actual depth of the liquid in the tank ? (AISSCE 1991) Calculate the refractive index of the material of an equilateral prism for which the angle of minimum deviation is 60º (AISSCE 1994) An object is kept at 0.2 m from a convex lens of focal length 0.15 m Find the position of the image produced (AISSCE 1993) The radii of curvature of a double convex lens are 15 cm and 30 cm It’s refractive index is 1.5 Calculate the focal length of the lens (AISSCE 1992) The radii of curvature of a double concave lens are 30 cm and 60 cm and its refractive index is 1.5 Calculate the focal length of the lens (AISSCE 1992) A compound microscope has a magnification of 30 The focal length of its eyepiece is cm Assuming the final image to be formed at the least distance of distinct vision, calculate the magnification produced by the objective (AISSCE 1991) Roughly calculate the limit of resolution of a 100 cm telescope with visible light of wavelength λ = 5500 Å Two thin lenses of focal lengths + 10 cm and – cm are kept in contact What is (a) the focal length (b) the power of the combination (AISSCE 1993) A ray of light passes through an equilateral glass prism, such that the angle of incidenc is equal to the angle of emergence If the angle of emergence is 3/4 times the angle of prism, calculate the refractive index of the glass prism (AISSCE Delhi 2000) 10 An object is placed at 0.06 m from a convex lens of focal length 0.10 m Calculate the position of the image (AISSCE 1993) S Chand & Company Limited 11 The focal length of the objective of an astronomical telescope in 75 cm and that of the eyepiece in cm If the final image is formed at the least distance of distinct vision from the eye, calculate the magnifying power of the telescope 12 Two lenses of powers - 2.25 D and + 1.75 D are kept in contact What is the effective focal length of the combination ? 13 Light strikes a glass plate at an angle of incidence 60º, part of the beam being reflected and part refracted It is observed that the reflected and the refracted rays are inclined to each other at an angle of 90º What is the refractive index of glass 14 A needle placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of lens Identify the type of lens and determine its focal length What is the size of the image, if the size of the object is 5.0 cm ? 15 A convex lens of focal length 0.2 m, made of glass of µ =3/2, is immersed in water (µ = 4/3) Find the change in the focal length of the lens 16 Two thin lenses of focal lengths 25 cm and 50 cm are kept in contact What is the power of the combination ? 17 Calculate the dispersive power of a crown glass prism if the angles of deviation for violet, yellow and red colours are, respectively, 4.3º, 4.0º and 3.7º 18 What is the magnifying power of a telescope whose objective and eyepiece have focal lengths 150 cm and cm respectively ? 19 A small telescope has an objective of focal length 140 cm and an eyepiece of focal length 5.0 cm What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment ? (b) the final image is formed at the least distance of distinct vision (25 cm) ? (NCERT Book) 20 The focal length of the objective and the eyepiece of a compound microscope are cm and cm respectively If an object is placed at a distance of cm from the objective, what is the magnification produced by the microscope ? Take D = 25 cm S Chand & Company Limited 21 A glass slab is immersed in water Find the critical angle at glass - water interface Given aµg = 1.5 and aµw = 1.33 22 A small pin fixed on a table top is viewed from above, from a distance of 50 cm By what distance would the pin appear to be raised, if it is viewed through a 15 cm thick glass slab held parallal to the table ? Refractive index of glass = 1.5 Does the answer depend on the location of the slab (NCERT Book) 23 Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature What is the radius of curvature required if the focal length is to be 20 cm ? (NCERT Books) 24 A glass lens has a focal length of cm in air What will be its focal length in water ? Refractive index of glass is 1.51 and that of water is 1.33 25 Show mathematically that the magnification produced by a concave lens is always less than unity 26 At what angle should a ray of light be incident on a face of a prism of refracting angle 60º so that it just suffers total reflection inside the prism? µ = 1.5 27 The refractive index of a prism of refracting angle 60º is 1.62 What is the angle of minimum deviation ? 28 The focal length of the objective of a telescope is 100 cm and the focal length of the eyepiece is cm What is the magnifying power of the telescope when the image is (a) at infinity (b) at the least distance of distinct vision (25 cm) ? 29 An astronomical telescope consists of two thin lenses set 36 cm apart, and has a magnifying power Calculate the focal lengths of the lenses (AISSCE Delhi 1997) 30 Light from a point source in air falls on a spherical (convex) glass surface (µ = 1.5, radius of curvature = 20 cm) The distance of the light source from the glass surface is 100 cm At what position is the image formed ? (NCERT Book) S Chand & Company Limited 31 The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm Its focal length is 12 cm What is the refractive index of glass (NCERT Book) 32 A beam of light converges to a point P A lens is placed in the path of the convergent beam 12 cm from P At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm (b) a concave lens of focal length 16 cm 33 At what angle should a ray of light be incident on the face of a prism of refracting angle 60º so that it just suffers total internal reflection at the other face ? The refractive index of the prism is 1.524 (NCERT Book) 34 Using a convex lens one obtains a sharp image of the slit on a screen The image size is measured to be 4.6 cm The lens is displaced away from the slit and at a certain location another sharp image of size 1.7 cm is obtained Determine the size of the slit (NCERT Book) ANSWERS µ = 1.3, Apparent depth = 7.7 cm µ= Real depth Apparent depth ∴ Real depth = 1.3 × 7.7 = 10.0 cm A = 60º, δm = 60º A + δm µ= sin ( A 2) sin µ= = sin (60º + 60º )  sin 60º = sin (60º 2) sin 30º S Chand & Company Limited u = – 0.2 m, or or or f = 0.15 m, v = ? 1 – = v u f 1 1 + = + = v f u 0.15 (–0.2) 4–3 = = 0.6 0.6 v v = 0.6m R1 = + 15 cm, R2 = – 30 cm, µ = 1.5, f = ?  1  = (µ –1)  –  f  R1 R2    = (1.5 –1)  +   15 30   +  0.5 × = (0.5)   = 30  30  10 or f = = 20cm 0.5 R1 = – 30 cm, R2 = + 60 cm, µ = 1.5, S Chand & Company Limited  1  = (µ –1)  –  f R2   R1  1  = (1.5 –1)  –  60   – 30   = 0.5 ×    20  20 = – 40cm or f=– 0.5 M = 30, fe = cm M = mo × me me =1 + D 25 =1+ =6 fe ⇒ 30 = mo × or mo = Limit of resolution of a telescope, dθ = 1.22 λ d d = 100 cm = 1m –10 λ = 5500 × 10 m 1.22 × 5500× 10 –10 = 67.1 × 10 – radians dθ = S Chand & Company Limited f1 = + 10 cm f2 = – cm 1 1 = + = + f f1 f 10 –5 1– –1 = 10 10 f = –10 cm = 1 = = – 10D f – (0.1) Here A = 60º A 3  δm = i + e – A =  A  – A = = 30º   ( A + δm ) sin sin 45 2 µ= = = = sin A sin 30º 12 10 u = – 0.06 m; f = 0.10 m 1 – = v u f 1 or – = 0.10 v – 0.06 P = 1 –10 – 40 = – = = 0.6 v 0.10 0.06 or v = – 40 = – 0.15m Thus the image is 0.15 m from the lens on the same side as the object or S Chand & Company Limited 11 fo = 75 cm, fe = cm  D 1+  fe   75  25  = 1+    = 90 Magnifying power m = fo fe 12 Here P1= 2.25 D, P2 = 1.75 D P= P1 + P2 = – 2.25 + 1.75 = 0.50 D f= –1 = m P 0.50 = – 2.0 m 13 i = 60º sin i We have µ = sin r Here i + r = 90º ⇒ r = 90º – 60º = 30º sin 60º ∴ µ= sin 30º = i Air glass r 12 = Fig B.23 S Chand & Company Limited 14 u = – 45 cm, v = 90 cm, y = 5.0 cm 1 – = v u f 1 1 +2 or = + = = f 90 45 90 90 or f = 30 cm Lens is convex since f is +ve y′ v Also m = = y u y′ 90 or = – 45 y ′ = – 10 cm Negative sign appears because the image is inverted 32 =9 a µ w = , f a = 20 cm, wµ g = 15 a µ g = , 43  1  = (µ –1)  Now –  f R R 2  ⇒ 3   1  =  – 1  –  f a    R1 R  9   =  – 1  –  fw    R1 R  Dividing Eq (1) by Eq (2) .(1) .(2) S Chand & Company Limited f w (3 – 1) = =4 fa (9 – 1) fw = fa Change in focal length = fw – fa = fa – fa =3fa = × 20 = 60 cm 16 f1 = 25 cm = 0.25 m, f2 = 50 cm = 0.50 m P1 = 1 = = 4D f1 0.25 P2 = 1 = = 2D f2 0.50 Power of the combination, P = P1 + P2 = 6D 17 δV = 4.3º, δY = 4.0º, δR = 3.7º δV −δ R Dispersive power ω = δY 4.3 – 3.7 0.6 = = 4.0 4.0 = 0.15 18 fo = 150 cm, fe = cm ∴ m= f o 150 = = 50 fe S Chand & Company Limited 19 fo = 140 cm, fe = 5.0 cm f o 140 = = 28 fe (b) When the final image is formed at the least distance of distinct vision, (a) When the telescope is in normal adjustment, m = fo  f  1+ e   fe  D 140   = 28 (1 + 5) = 1+  25  = 28 × = 33.6 m = 20 fo = cm, fe = cm D = 25 cm Distance of object from objective, uo = – cm 1 – = vo uo fo or 1 1 1 3– – = or = – = = –6 4 12 12 vo vo or vo =12 cm Now  D  12  25  1+  = 1 +  f   e   12 31 31 = × = = 10.33 6 m= vo uo S Chand & Company Limited 21 µ = 1.5, aµw = 1.33 1.5 aµg = = 1.13 wµ g = µ 1.33 a w a g Now, or wµ g = or sin ic = sin ic 1 = = 0.8867 1.13 wµ g ic = 62º 28′ 22 Real depth = 15 cm µ = 1.5 Real depth 15 = = 10 cm µ 1.5 The pin appears to be raised by a distance = 15 – 10 = 5.0 cm dependent of the location of the slab (for small angles of incidence) Apparent depth = 23 µ = 1.55, R1 = R, R2 = –R, f = 20 cm or or  1  – = (µ –1)   f R2   R1 1 1 = (1.55 –1)  +  20 R R = (0.55) × 20 R R = 0.55 × × 20 = 22.0 cm S Chand & Company Limited The answser is in- 24 Here fa = cm µ = 1.51 a g µ = 1.33 a w  1  – = ( a µ g –1)   fa R R    1  – = ( w µ g – 1)   fw R R   ⇒ or    µ –1  a g  =   µ µ – a g w g –1  a w 1.51 ì5 fw = 1.51   1.33    fw = fa a g (1.51 1) ì ì1.33 (1.51 1.33) = 18.8cm 1 25 Using the new cartesian sign convention the lens equation is – = v u f Now, u = –| u | For a concave lens f = – | f| and v = – | v| = S Chand & Company Limited So ⇒ 1 1 1 – = or = + – |v | –| u | – | f | |v | | u | | f | 1 > or | v | < | u | |v| |u| v