Physics-25.19 -Ray Optics

1. SOME IMPORTANT DEFINITIONS USED IN OPTICS We define a varied collection of useful terms. Some of these definitions will be revisited when the appropriate topics are taken up. (i) Ray: The path of light, as determined within the approximations of geometric optics, is a ray. In a homogeneous medium, it is a straight line. (ii) Beam: A collection of rays, usually referred to as a bundle of rays, forms a beam. One may have a convergent, divergent or, a parallel beam. A convergent beam may converge to a point or, a line, a divergent beam may diverge from a point or, a line. A parallel beam consists of parallel rays. (iii) Collimation: A process whereby a divergent (or, convergent) beam is rendered parallel usually through the use of lenses and/or mirrors. (iv) Object & Image: The term object is used to refer to any object (being photographed or observed) that is a source of light, and its likeness (usually two dimensional) formed or observed by an optical system is the image. (v) Optical System: An optical system consists of elements like lenses, mirrors, prisms, etc. (vi) Axis: The axis of an optical system is frequently an axis of symmetry such that a ray directed along the axis continues in the same direction or, returns backwards (if reflected within the system) (vii) Centre of Curvature: Most lenses and curved mirrors being manufactured spherical (i.e., their surfaces are spherical), the centre of curvature of the curved mirror or, the curved surface of a lens is important and is frequently denoted by the letter C. The radius of curvature is also equally important in the analysis. (viii) Pole: The pole of a spherical surface (refracting or, reflecting) is the central point of the surface involved in the formation of the image. It is denoted by O or, P. The axis, for a single spherical surface, is the join of the pole with the centre of curvature; it is known as the principal axis. (ix) Optical Centre: The optical centre of a thin lens is a point on the axis of the lens such at a ray directed towards that point emerges parallel to itself after passing through the lens. (x) Paraxial Rays: It is observed that the formation of clear images by spherical surfaces takes place only with rays which are close to the principal axis and make very small angles with it. (xi) Wave speed: It is the distance travelled by the wave disturbance in a unit time. It is denoted by the letter . (xii) Frequency: It is the number of vibrations made by the medium particle in 1 s. In other words, it is number of waves passing through a point of the medium in 1 second. It is generally represented by the letter n or . Its unit is hertz and it is represented as Hz or s–1. (xiii) Wavelength: It is the distance travelled by the wave in one complete period of a medium particle. In other words, distance between two consecutive crest or trough. Relationship between the wavelength, wave speed and frequency: If the wave speed is and period is T, then by definition Wavelength  Distance travelled in one period  Distance travelled in T s  x T or But Or i.e., Wave speed  Frequency X Wavelength. 2. REFLECTION AND REFRACTION AT PLANE AND SPHERICAL SURFACES 2.1 REFLECTION AT PLANE SURFACE (i) Reflection of Light When light rays strike the boundary of two media such as air and glass, a part of light is turned back into the same medium. This is called Reflection of Light. The wavelength and the velocity of the light wave remains the same. Laws of reflection are obeyed at every reflecting surface, i.e. (ii) Laws of Reflection (a) The incident ray (AB), the reflected ray (BC) and normal (BN) to the surface (SS') of reflection at the point of incidence (B) lie in the same plane. This plane is called the plane of incidence (or plane of reflection). (b) The angle of incidence (the angle between normal and between the reflected ray and the normal are equal, i.e. (iii) Types of reflection (a) Regular Reflection (b) Diffused Reflection or Scattering or Diffusion (a) Regular Reflection: When the reflection takes place from a perfect plane surface it is called Regular Reflection. (b) Diffused Reflection or Scattering or Diffusion: When the surface is rough light is reflected from the surface from bits of its plane surfaces in irregular directions. This is called diffusion. This process enables us to see an object from any position. Characteristics of Reflection by a Plane Mirror (i) Distance of object from mirror = Distance of image from the mirror. (ii) The line joining the object point with its image is normal to the reflecting surface. (iii) The image is laterally inverted (left right inversion). (iv) The size of the image is the same as that of the object. (v) For a real object the image is virtual and for a virtual object the image is real. (vi) The minimum size of a plane mirror, required to see the full self image, is half the size of that person. (vii) For a light ray incident at an angle 'i' after reflection angle of deviation (viii) If i = 0 then r = 0, this implies that a ray of light incident normally on a mirror retraces its path. (ix) The eye always observes an object in the direction in which the rays enter the eye (x) The laws of reflection holds good for all kinds of reflection. (xi) Image of an object is the point at which rays after reflection (or reflection) actually converge or appear to diverge from that point. (xii) If the direction of the incident ray is kept constant and the mirror is rotated through an angle  about an axis in the plane mirror then the reflected ray rotates through an angle 2. (xiii) If an object moves towards (or away from) a plane mirror at a speed v, the image will also approach (or recede) at the same speed. Further the relative speed of image to the object will be v – (–v) = 2v. (xiv) If two plane mirror are inclined to each other at 90, the emergent ray is always antiparallel to incident ray, if reflected from each mirror, irrespective of angle of incident. (xv) When two plane mirror, inclined to each other at an angle , the number of images formed can be determined as follows: (a) If is an even integer, say ‘p’, then number of image formed say q = p – 1, for all position. (b) If is an odd integer say ‘q’ then number of image formed say p = q, if the object is not on the bisector of the angle between mirrors. Also, p = q – 1, if the object is on the bisector. (c) If is a fraction, then the number of image formed will be equal to its integral part. (xvi) The images are laterally inverted. (xvii) The linear magnification is unity. Illustration 1: A ray of light on a plane mirror along a vector . The normal on incident point is along . Find a unit vector along the reflected ray. Solution: Reflection of a ray of light is just like an elastic collision of a ball with a horizontal ground. Component of incident ray along the inside normal gets reversed while the component perpendicular to it remains unchanged. Thus the component of incident ray vector parallel to normal, i.e., gets reversed while perpendicular to it, i.e., remains unchanged. Thus, the reflected ray can be written as A unit vector along the reflected ray will be  = 2.2 REFLECTION-AT SPHERICAL MIRRORS A spherical mirror is a reflecting surface which forms a part of a sphere (as shown in following a and b diagram). When the reflection takes place from the inner surface and outer surface is polished or silvered the mirror is known as concave mirror. Vice- versa, it is convex. Ray diagram for concave and convex mirror 2.3 CHARACTERISTICS OF REFLECTION BY A SPHERICAL REFLECTING SURFACES (FOR SMALL APERTURE) (i) Pole (P) is generally taken as the mid point of reflecting surface. (ii) Centre of curvature (C) is the centre of the sphere of which the mirror is a part. (iii) Radius of curvature is the radius of the sphere of which the mirror is a part. Distance between P and C. (iv) Principal Axis is the straight line connecting pole P and centre of curvature C. (v) Principal focus (F) is the point of intersection of all the reflected rays which strike the mirror (with small aperture) parallel to the principal axis. In concave mirror it is real and in the convex mirror it is virtual. (vi) Focal length (f) is the distance from pole to focus. (vii) Aperture is the diameter of the mirror. (viii) If the incident ray is parallel to the principal axis, the reflected ray passes through the focus. (Fig (a)) (ix) If the incident ray passes through the focus, then the reflected ray is parallel to the principal axis (Fig.(b)) (x) Incident ray passing through centre of curvature will be reflected back through the centre of curvature. (Fig.(c)) (xi) where f = focal length R = Radius of curvature. (xii) Sign convention and magnification. We follow Cartesian coordinate system convention according to which: (a) The pole of the mirror is the origin (b) The direction of the incident rays is considered as positive x-axis. (c) Vertically up is positive y-axis Note: Radius of Curvature and Focal Length of: (a) Concave mirror is positive (b) Convex mirror is positive (xiii) (a) Linear Magnification or lateral magnification or transverse magnification coordinate of image; coordinate of the object (both perpendicular to the principle axis of mirror) (b) Longitudinal Magnification (for any size of the object) (for short object) Tracing for spherical mirror Concave Convex (1) A ray going through centre of curvature is reflected back along the same direction. (2) A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays after reflection intersect on the focal plane. (3) A ray going to the pole and the reflected ray from it make equal angles with the principal axis. 2.4 REAL AND VIRTUAL SPACES A mirror, plane or spherical divides the space into two: (i) A side, where the reflected rays exist. This is real space. (ii) The other side where the reflected rays do not exist. This is virtual space Object Object is decided by incident rays only. The point object for mirror is that point. (a) From which the rays actually diverge to be incident on the mirror (Real object). Or (b) Towards which the incident rays appear to converge (virtual object). Image Image is decided by reflected or refracted rays only. The point image for a mirror is that point. (a) Towards which the rays reflected from the mirror, actually converge (real image). Or (b) From which the reflected rays appear to diverge (virtual image). NEW CARTESIAN SIGN CONVENTION – FOR MIRRORS AND LENSES There are several sign-conventions. Students should try to follow any one of these. In this package, the new Cartesian sign convention has been used wherever required. (i) All distance are measured from the pole. (ii) Distances in the direction of incident rays are taken as positive. (iii) Distances in the direction of incident rays are taken as negative. (iv) Distances above the principal axis are taken as positive. (v) Distances below the principal axis are taken as negative. (vi) Angles measured from the normal, in anti-clockwise direction are positive, while in clockwise direction are negative. 2.5 MIRROR FORMULA Consider the shown figure where O is a point object and I is corresponding image. CB is normal to the mirror at B. By laws of reflection, OBC = PBC =   +  = ,  +  =    +  = 2 For small aperture of the mirror, , ,   0    tan ,   tan ,   tan , P’  P  tan  + tan  = 2 tan  Applying sign convention, u = -OP, v = -IP, R = -CP If u = , , but by definition, if u = , v = f. Hence, For convex mirrors, an exactly similar formula emerges. Illustration 2: An object is placed 21cm infront of a concave mirror of radius of curvature 10cm. A glass slab of thickness 3cm and refractive index 1.5 is then placed closed to the mirror in the space between the object and mirror. Find the position of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1cm) Solution: Given that radius of curvature = 10 cm  focal length f = 5 cm Here Where 1.5 = refractive index of slab Using the formula for concave mirror Solving we get Applying the correction due to the slab, we have final image distance = |v| + 1 2.6 REFRACTION AT A PLANE SURFACE Laws of Refraction (Snell’s Law) (i) The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane. (ii) The ratio of the sines of the angle of incidence (i) and of the angle of refraction (r) is a constant quantity  for two given media, which is called the refractive index of the second medium with respect to the first. When light propagates through a series of layers of different medium as shown in the figure, then the Snell’s law may be written as 1 sin 1 = 2 sin 2 = 3 sin 3 = 4 sin 4 = constant In general,  sin  = constant Fig. A series of transparent layers of different refractive indices Fig. A light ray passing from air to water bends toward the normal When light passes from rarer to denser medium it bends toward the normal as shown in the fig. According to Snell’s law 1 sin 1 = 2 sin 2 When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig. Fig. A light ray passing from water to air bends away from the normal. For a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams. Determination of the image formed for the cases given becomes much simpler when we restrict ourselves to nearly normal incident rays. CASE – I (Object in the denser medium) For nearly normally incident rays 1 and 2 will be very small. Similarly, The same result is obtained for the other case also. The image distance from the refracting surface is also known as Apparent depth or height. Apparent Shift Apparent shift = Object distance from refracting surface – image distance from refracting surface. y (apparent shift) where t is the object distance and • If there are a number of slabs with different refractive indices placed between the observer and the object. Total apparent shift = yi Illustration 3: A person looking through a telescope T just sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid ( = 1.5), he observes a mark at the centre, B, of the vessel. What is the height of the vessel if the diameter of its cross-section is 10cm? Solution: It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same setting; this means that the images of point B, is observed at A, because of refraction of the ray at C. For refraction at C, Now sin Where h is the height of vessel. 2.7 REFRACTION AT SPHERICAL SURFACE When light strikes a spherical (curved) surface, the position of object (O), the image (I) and the radius of curvature (R) are related as: ; here 2 > 1 Consider another case, a spherical surface of radius R separating two media with refractive indices . The radius of curative R will be taken positive when incident rays strike the convex side of the surface. If u is the object distance and v is the image distance, For light rays going from medium 1 (1) to medium 2 (2): The corresponding equation for refraction at a plane surface will be: (taking R = ) Illustration 4: A transparent rod 40 cm long is cut flat at one end and rounded to a hemispherical surface of 12 cm radius at the other end. A small object is embedded with in the rod along its axis and half way between its ends. When viewed from the flat end of the rod, the object appears 12.5 cm deep. What is its apparent depth when viewed from the curved end? Solution: For the flat surface: Real depth of the object = 20 cm Apparent depth = 12.5 cm Using For the curved surface: We will use u = 20 cm, R = –12 cm, Hence the object appears 33.3 cm deep from the curved side. 3. CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION Consider a ray of light that travels from a denser medium to a rarer medium. The angle of incidence for which the angle of refraction becomes 90 is called critical angle. This is used for constructing totally reflecting prisms. and if i > c When the angle of incidence of a ray travelling from denser to rarer medium is greater then the critical angle, no refraction occurs. The incident ray is totally reflected back into the same medium. Here the laws of reflection hold good. Some light is also reflected before the critical angle is achieved, but not totally. Illustration 5: A point source of light a placed at the bottom of a tank containing a liquid of refractive index  The level of the liquid is at a height h above the bottom of the tank. A bright circular spot is seen on the surface of the liquid when viewed from above. Find the radius of the spot. Solution: The rays of light emerging from the source, which are incident at an angle . (where is the critical angle) will be reflected. Further, or, 4. THIN LENS Consider the thin lens shown here with the two refracting surfaces having radii of curvature equal to R1 and R2 respectively. The refractive indices of this surrounding medium and of the material of the lens are 1 and 2 respectively. Now using the result that we obtained for refraction at single spherical surface we get, For first surface, ...(1) For second surface, ...(2) Adding (1) and (2), When u = , v = f, Lensmaker's Formula Also, Thin Lens Formula EQUIVALENT FOCAL LENGTH OF TWO OR MORE THIN LENSES IN CONTACT Special cases (a) If the lens is immersed in a liquid whose refractive index is greater than refractive index of material of the lens then fliquid is the focal length of lens in the liquid Ig is the R.1. of glass w.r.t. liquid. The focal length of the lens in the liquid becomes negative. (b) If the lens is immersed in a liquid whose R.1. is equal to the R.1. of the material of the lens then  The focal length becomes infinite (c) If the lens be immersed in a liquid whose R.1. is less than the R.1. of the material of the lens The focal length of the lens increases. In general, if  decreases, f increases Note: A lens is converging if its focal length is positive and diverging if focal length is negative (in neo-Cartesian sign convention). From this we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging. Ray tracing for lens Convex Concave (i) A ray passing through optic centre goes undeviated. (ii) When a ray is incident parallel to principal axis, the refracted ray passes or appears to pass through the focus. Illustration 6: A glass rod has ends as shown in figure, The refractive index of glass is . The object point O is at a distance 2R from the surface of larger radius of curvature. The distance between the apexes of the ends is 3R. Show that the image point of O is formed at a distance of Solution: From the right hand vertex. For the first surface Here 2 =, 1 =1,  = –2R, R1 = +R For the second surface Here 2 = 1, 1 = , 5. MAGNIFICATION (i) Magnification in Mirror In new Cartesian sign convention, we define magnification is such a way that a negative sign (of m) implies inverted image and vice-versa. A real image is always inverted one and a virtual one is always erect. Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirror and on opposite sides in case of lenses, we define m as in case of reflection by spherical mirror as: (ii) Lateral Magnification for Refracting Spherical Surface Hence, (iii) Magnification of A Thin Lens Illustration 7: An object is placed at a distance of 10 cm to the left on the axis of a convex lens A of focal length 20 cm. A second convex lens of focal length 10 cm is placed co-axially to the right of the lens A at a distance of 5 cm from A. Find the position of the final image and its magnification. Trace the path of the rays. Solution: Here for 1st lens, u1 = –10 cm f1 = 20 cm. i.e. the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens. For 2nd lens Here i.e. the final image is at a distance of on the right of the second lens. The magnification of the image is given by, 6. POWER OF A LENS If f is in metres then the power P of the lens in dioptres is given by, If two lens in are separated by a distance d then power of combination of lens is Where P1 and P2 are optical powers of the two lenses, and  is the refractive index of the medium in between them. Illustration 8: A lens has a power of +5 diopters in air. What will be its power if completely immersed in water? Solution: Let fa and fw be the focal lengths of the lens in air and water respectively, then fa = 0.2 m = 20 cm Now ...(1) and ...(2) Dividing equation (1) by equation (2), we get, Again, 7. SILVERING AT ONE SURFACE OF LENS When one surface of a thin lens is silvered, then the focal length F of the effective lens-mirror combination is expressed as, , where fi is the focal length of the lens or mirror to be repeated as many times as the refraction or reflection respectively is repeated. Some Cases: (i) Focal length of plano–convex lens when silvered at its plane surface When an object is placed in front of such a lens. The ray first of all are refracted from the convex surface, then reflected from the polished plane surface and again refracted out from the convex surface. If and fm be the focal lengths of lens and mirror. (ii) Focal length of plano –convex lens when silvered at convex surface (iii) Focal length of convex lens where convex surface of radius R2 is silvered Illustration 9: An object is 1 metre in front of the curved surface of a plano–convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens? Solution: Here u = 100 cm v = 120 cm Now Where FL – focal length of the lens. FM – focal length of the mirror. 8. ANGLE OF DEVIATION OF A RAY WHEN IT PASSES A LENS O is the object and  is the image  is the angle of deviation. 9. DETERMINATION OF FOCAL LENGTH OF A CONCAVE MIRROR BY U–V METHOD The relation between object distance u and the image v from the pole of the mirror is given by, Where f is the focal length of the mirror. The focal length of the concave mirror is obtained either from versus graph or from u–v graph. and graph When the image is real (of course only upon then it can be obtained on screen), the object lies between focus (F) and infinity. In such a situation u, v and f all are negative. Hence the mirror formula Becomes, or again, or, Comparing with y = mx + c, the desired graph will be a straight line with slope –1 and intercept equal to The corresponding versus graph is as shown in figure. The intercepts on the horizontal and vertical axes are equal. It is equal to . A straight line OC at an angle 45o with the horizontal axis intersects line AB at C. The coordinates of point C are The focal length of the mirror can be calculated by measuring the coordinates of either of the points A, B or C. u–v graph The u–v graph comes out to be a hyperbola as shown in figure. A line drawn at angle 45o from the origin intersects the hyperbola at point C. The coordinates of point C are (2f, 2f). The focal length of mirror can be calculated by measuring the coordinates of point C. Determination of focal length of convex lens using u–v method The relation between u, v and f for a convex lens is, and graph Using the proper sign convention, u is negative, u negative, v and f are positive. So we have, or Comparing with y = mx + c, graph between and is negative with slope –1 and intercept The corresponding graph is as shown in figure. Proceeding in the similar manner as discussed in case of a concave mirror the focal length of the lens can be calculated by measuring the coordinates of either of the points A, B and C. u–v Graph The u versus v graph is a hyperbola as shown in figure. By measuring the coordinates of point C whose coordinates are (2f, 2f) we can calculate the focal length of the lens. 10. PRISM A prism has two plane surfaces AB and AC inclined to each other as shown in following figure. A is called the angle of prism or refracting angle. (i) Refraction Through A A light ray striking at one face of a triangular glass prism gets refracted twice and emerges out from the other face as shown above. The angle between the emergent and the incident rays is called the angle of deviation (D). The angle between the two refracting faces involved is called the refracting angle (A) of the prism. From AXY, we have: A + (90° – r1) + (90° – r2) = 180° (i) Deviation D = (i – r1) + (e – r2) D = (i + e) – (r1 + r2) D = i + e – A A + D = i + e (ii) We also have two equations from Snell's Law at X & Y. (i) Angle of Deviation It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angles of incidence and emergence get interchanged but the angle of deviation remains same. angle of incidence = 1 angle of emergence = 2 angle of incidence = 2 angle of emergence = 1 Hence the same angle of deviation D is possible for two different angles of incidence : 1 and 2, where 1 + 2 = A + D. (ii) Minimum Deviation The angle of deviation is minimum when the path of light ray through the prism is symmetrical. i.e., angle of incidence = angle of emergence i = e and hence also from we have: (iii) Grazing Incidence When i = 90°, the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass - air. This is known as grazing incidence. (iv) Grazing Emergence When e = 90°, the emergent ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for glass - air. This is known as grazing emergence. (v) Maximum Deviation The angle of deviation is same for both the above cases (grazing incidence & grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. (vi) Dispersion of Light When a ray of light passes through a prism, it splits up into rays of constituent colours or wavelengths. This phenomenon is called dispersion of light. The refractive index of a medium is different for light rays of different wavelengths. Larger the wavelengths, the lesser is the refractive index. i.e., A ray of white light passing through a prism gets splits into different colours because the deviation is more for violet and less for red. This phenomenon is known as dispersion. For a prism with very small refracting angle A, the deviation D is given by: Hence deviation of violet = Dv = (v, ¬¬– 1)A and for red = Dr = (r – 1)A The angle between the red and violet rays is known as the angle of dispersion. The angle of dispersion Angle of mean deviation Dispersive power of glass This phenomenon arises due to the fact that refractive index varies with wavelength. It has been observed for a prism that  decreases with the increase of wavelength, i.e. . Angular dispersion, (vii) Dispersive power where  is deviation of mean ray (yellow) As (viii) Deviation without Dispersion This mean an achromatic combination of two prisms in which net or resultant dispersion is zero and deviation is produced. For the two prisms, Where  and ' are the dispersive powers of the two prisms and  and ' their mean deviations. (ix) Dispersion without deviation A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism is called a direct vision prism. This combination produces dispersion without deviation. For deviation to be zero, ( + ') = 0 Illustration 10: A prism of crown glass refracting angle of 50 and mean refractive index = 1.51 is combined with one flint glass prism of refractive index = 1.65 to produce no deviation. Find the angle of flint glass and net dispersion. Given: v = 1.523, R = 1.513 (for crown glass) 'v = 1.665, 'R = 1.645 (for flint glass) Solution: Let A' be the angle of flint glass prism. Here A = 5 and  = 1.51 for crown glass prism. Deviation produced by flint glass For no deviation ' =  or 0.65 A' = 2.55 Net dispersion, 11. ASSIGNMENT 1. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 30 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of time the ray undergoes reflections (including the first one) before it emerges out is: (A) 28 (B) 30 (C) 32 (D) 34 Solution: (B) No. of ref 2. Consider the metal coin in figure. The right half is polished, while left half is unpolished. In a dark room, this coin is placed face up on the floor and illuminated with a thin-beamed flashlight, as shown in figure. A and B look at the coin from the position as indicated in figure According to A, which half of the coin (if either) is brighter? (A) The polished half (B) The unpolished half (C) Both halves look equally bright (D) Both halves look completely dark Solution: (A) The light reflects of the coin and reaches A’s eye. As the polished half is smoother than the unpolished half, hence the polished half behaves more like a mirror. So, most of the light striking the polished half of the coin reflects instead of going diffusion reflection. Hence, A sees a high percentage of the light that hits the polished part of the coin. Light hitting the unpolished half undergoes more diffuse reflection, bouncing off the coin in all different directions. Hence, a smaller percentage of that light reaches A’s eyes. So, for A the polished half looks more brighter than the unpolished half. 3. A glass prism of refractive index 1.5 is immersed in water (R.I. = 4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if (A) sin  8/9 (B) sin  < 2/3 (C) 2/3 < sin  < 8/9 (D) None of these Solution: (A) The light ray is totally reflected internally at the point P (say). Therefore the critical angle C  . Also,  sin c =  sin  8/9 4. A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing: (A) a convex mirror of suitable focal length (B) a concave mirror of suitable focal length (C) a convex lens of focal length less than 0.25 m (D) a concave lens of suitable focal length. Solution: (C) mage can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens. Let u = 2f + x, then It is given that u + v = 1m. or, or, This will be true only when f < 0.25 m. 5. A screen beaming a real image of magnification m1 formed by a convex lens is moved through a distance x. The object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is: (A) (B) (C) (D) None of these Solution: (A) In first case, =m1  1+m1 = …(1) In the second case And  m2 = …(2) (1) and (2)  m2 – m1 = x/f  f = 6. If incident ray MP and reflected ray QN are parallel to each other, find the angle  between the mirrors. (A) 60 (B) 45 (C) 90 (D) 180 Solution: (C) The deviation produced by first mirror = 180 – 2i1 and the deviation produced by second mirror = 180 – 2i2. The total deviation is 180 as MP and QN are anti parallel So, 180 = 360 – 2(i1 + i2)  i1 + i2 =  = 90 7. A 16 cm long image of an object is formed by a convex lens on a screen placed normal to its principal axis. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is: (A) 9 cm (B) 11 cm (C) 12 cm (D) 13 cm. Solution: (C) 8. A small mirror of area ‘A’ and mass ‘m’ is suspended in a vertical plane by means of a weightless string. A beam of light of intensity ‘I’ falls normally on the mirror and the string is deflected form the vertical through a very small angle ‘’. Assuming the mirror to be perfectly reflecting, obtain an expression for . (A) (B) (C) (D) none of the above Solution: (A) When the light is incident on the mirror is totally reflected and there is no loss of energy.  Change in momentum = 2  incident momentum Intensity = Momentum = Pressure, P = = = (where c = speed of light)  = tan = = 9. If x and y be the distances of the object and image formed by a concave mirror from its focus and f be the focal length then (A) xf = y2 (B) xy = f2 (C) x/y = f (D) x/y = f2 Solution: (B) According to Newton's formula xy = f2 . Note that m = or  xy = f2 10. An observer can see, through a pin-hole, the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is: (A) 5 / 2 (B) (C) (D) 3 / 2. Solution: (B) ...(i) ...(ii) Using (i) and (ii), 11. The sun (diameter d) subtends an angle  radians at the pole of a concave mirror of focal length f. What is the diameter of the image of the sun formed by the mirror? (A) f (B) f/2 (C) f2 (D) f2 Solution: (A) Since the sun is at a very large distance, u is very large and so (I/u) is practically zero. So, i.e., v = f i.e., the image of sun will be formed at the focus and will be real, inverted and diminished. Now as the rays from the sun subtend an angle  radians at the pole, then in accordance with figure.  = [where d is the diameter of the image of the sun] i.e. d = f 12. A ray of light is incident on the left vertical face of a glass cube of refractive index 2, as shown in figure. The plane of incident is the plane of the page, and the cube is surrounded by liquid (1). What is the largest angle of incidence 1 for which total internal reflection occurs at the top surface? (A) sin1 = (B) sin1 = (C) sin1 = (D) sin1 = Solution: (A) Consider A Point Applying Snell’s law, 1sin1 = 2sin2 …(1) But 2 = 90 – c  cos2 = sinc = …(2) Elimination of 2 between (1) and (2), we get  sin1 = 13. A ray incident at a point as an angle of incidence of 60° enters a glass sphere of R.I. n = and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is: (A) 50° (B) 60° (C) 90° (D) 40° Solution: (C) Refraction at P.   r1= 30° Since r2 = r1  r2 = 30° Refraction at Q Putting R2= 30° we obtain i2 = 60° Reflection at Q  = 180°- (30°+60°) = 90° 14. What is the number of images of an object formed by two mirrors if the angle between the mirrors is in between 2/4 and 2/5? (A) 10 (B) 8 (C) 6 (D) 4 Solution: (D) As we know, if angle between two mirrors is between and then the no. of images formed is ‘n’ So here in the question it is and  no. of images will be 4. 15. If one face of a prism of prism angle 30 and  = is silvered, the incident ray retraces its initial path. What is the angle of incident? (A) 90 (B) 180 (C) 45 (D) 55 Solution: (C) As in AED 30 + 90 + D = 180 D = 60 Now as by construction D + r = 90 r = 90 – 60 = 30 Applying Snell’s law at surface AC sin i = ( )sin30 =  = i = sin–1 = 45 16. Two plane mirrors are inclined to each other at angle . A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray (A) 360 – 2 (B) 360 + 2 (C) 180 – 2 (D) 180 + 2 Solution (A) A ray AB is incident on mirror OM1 at angle  and is reflected along BC suffering a deviation 1 = FBC The ray BC falls on mirror OM2 at an angle of incidence  and is reflected along CD suffering another deviation 2 = GCD The total deviation is  = 1 + 2 It is clear from the diagram that 1 = 180 - 2 and 2 = 180 - 2   = 1 + 2 = 360 –2( + ) Now, in triangle OBC, OBC + BCO + BOC = 180 or (90 – ) + (90 – ) +  = 180 or  +  =  Hence  = 360 – 2 Which is independent of the angle of incidence  at the first mirror. 17. A layer of oil 3cm thick is floating on a layer of coloured water 5cm thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7cm. Find the refractive index of oil. (A) 1.6 (B) 1.4 (C) 1.9 (D) 0.9 Solution (B) Apparent depth (A) 18. A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by first surface of sphere (A) 24 cm (B) 30 cm (C) 34 cm (D) 50 cm 19. A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by second surface of sphere (A) 4 cm (B) 6 cm (C) 15 cm (D) 50 cm Solution (18(A); 19(A)) As the sun is very far away, the incident ray can be assumed as coming from infinity. For refraction at first surface: Using . . . [18 (A)] The rays try to converge at 1 and hence for the second surface it is a virtual object. For second surface: u = –(24 – 16) = –8 cm [19 (A)] Hence the final image is formed at 2 which is 4 cm from the other surface of the sphere. 20. If the critical angle is 4842’ when the media concerned are air and water and 3647’ when they are in air and glass, what is it when they are in water and glass. (A) 5252' (B) 6252' (C) 6262' (D) 7070' Solution (A) We know that Similarly Now for glass and water media sin C = refractive index of water w.r.t. glass 21. Find the size of the image formed in the situation shown in figure. (A) 0.9 cm (B) 0.6 cm (C) 1.2 cm (D) 1.4 cm Solution (B) Here u = –40 cm, R = –20 cm  = 1, 2 = 1.33 We have, v = –32 cm. The magnification is The image is erect. 22. A convergent lens of 6 diopters is combined with a diverging lens of -2 diopters. Find the power of combination? (A) 4 diopter (B) 6 diopter (C) 8 diopter (D) 10 diopter Solution (A) Here P1 = 6 diopters, P2 = -2 diopters Using the formula P = P1 + P2 = 6 – 2 = 4 diopters 23. Select a graph between ‘v’ and ‘u’ for a concave mirror. (A) (B) (C) (D) Solution (A) For a concave mirror  Curve would be a hyperbola for spherical mirror. 24. A prism is made up of flint glass whose dispersive power is 0.053. Find the angle of dispersion if the mean refractive index of flint glass is 1.68 and the refracting angle of prism is 3. (A) 20.08 (B) 10.08 (C) 0.208 (D) 0.108 Solution (D) As the refracting angle is small deviation = ( – 1) A angle of dispersion = CMP: The following figure shows a simple version of a zoom The converging lens has focal length f1’ and the diverging lens has focal length f2 = –|f2|. The two lenses are separated by a variable distance d that is always less than f1’ also the magnitude of the focal length of the diverging lens satisfies the inequality |f2| > (f1 –d). If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left to the diverging lens, they will eventually expand to the original radius r0 at the same point Q. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius r0 entering the converging lens. 25. At the point where rays enters the diverging lens, the radius of the ray bundle decreases to: (A) r’0 = r0 (B) r0 = r’0 (C) r’0 = r0 (D) r’0 = r0 Solution: (A) From the similar triangles ABC and DEC so r’0 26. To the right of the diverging lens the final image I’ is formed at a distance s’2. (A) (B) (C) (D) Solution: (B) The image at focal point of the first lens, a distance f1 to the right of first lens, serves as the object for the second lens. The image is at distance f1–d to the second lens so s2 = –(f1 –d) = d – f1 so s’2 = 27. Find the effective focal length? (A) (B) (C) (D) Solution: (A) From the similar triangles in the sketch so f = 28. If f1 = 12 cm, f2 = – 18 cm find the maximum focal length of the combination? (A) 36 cm (B) 34 cm (C) 32 cm (D) 30 cm Solution: (A) Put the numerical values in the expression in the earlier question and we get 36 cm. 29. In continuation of fourth questions determine the minimum focal length of the combination (A) 21.6 cm (B) 24 cm (C) 30 cm (D) 35 cm Solution: (A) Put the numerical values in the above expression which we found and we will get 21.6 cm. 30. What value of d gives f = 30 cm? (A) 12 cm (B) 10 cm (C) 1.2 cm (D) 14 cm Solution: (C) f = 30 cm then 30 = 6 + d = 7.2 hence d = 1.2 cm. For test 1. An achromatic combination of two prisms - one of quartz and other of flint glass is to be formed. If the refracting angle of the quartz prism is 4, find the refracting angle of the flint glass prism. Also find the net mean deviation produced by the combination. Given that for the flint glass : v = 1.694, r = 1.65 and for the quartz glass : v = 1.557, r = 1.542. (A) –1.282 (B) +1.282 (C) –5.282 (D) –8.282 Solution (A) For an achromatic combination, net dispersion = 0 dispersion by one prism = dispersion by other prism. Net mean deviation 2. Find the dispersion produced by a thin prism of 18 having refractive index for red light = 1.56 and for violet light = 1.68. (A) 4.16 (B) 6.16 (C) 2.16 (D) 8.16 Solution (C) We know that dispersion produced by a thin prism Here 3. Calculate the dispersive power for crown glass from the given data (A) 0.1639 (B) 0.01639 (C) 1.639 (D) 2.639 Solution (B) Here Mean refractive index Dispersive power  is given by, 4. Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirrors. (A) 60 (B) 30 (C) 90 (D) 180 Solution: (A) Let  be the angle between the two mirrors OM1 and OM2. The incident ray AB is parallel to mirror OM2 and strikes the mirror OM1 at an angle of incidence equal to . From figure we have M1BA = OBC = M1OM2 =  Similarly for reflection at mirror OM2, we have M2CD = BCO = M2OM1 =  Now in triangle OBC, 3 = 180, therefore  = 60 5. A small filament is at the centre of a hollow glass sphere of inner and outer radii 8 cm and 9 cm respectively. The refractive index of glass is 1.50. Calculate the position of the image of the filament when viewed from outside the sphere. (A) 9 cm (B) – 9 cm (C) – 19 cm (D) +19 cm Solution: (B) For refraction at the first surface, u = –8 cm, R1 = –8 cm, 1 = 1, 2 = 1.5 v' = –8 cm It means due to the first surface the image is formed at the centre. For the second surface u = –9 cm, 1 = 1.5, R2 = –9 cm v = –9 cm Thus, the final image is formed at the centre of the sphere. 6. A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f. Such that its image which is real and elongated just touches the rod. Calculate the magnification. (A) 3/2 (B) 5/2 (C) 1/2 (D) 2/3 Solution: (A) Let be the length of the image. Then, Also image of one end coincides with the object, . Putting in mirror formula, 7. An object of length 2.5 cm is placed at a 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. Find the length of the image. Is the image erect or inverted? (A) – 8 cm (B) – 10 cm (C) – 5 cm (D) – 12 cm Solution: (C) The focal length F = –f and u = –1.5 f, we have, v = –3f Now, The image is 5 cm long. The minus sign shows that it is inverted. 8. Select a graph for concave mirror between 1/v and 1/u. (A) (B) (C) (D) Solution: (C) For a mirror let y = , x =  x + y = , y = – x + 1/f This is the equation of a straight line with slope = (–1) and intercept = Hence the graph would be 9. Show with the help of ray diagrams that a right angled isosceles prism can produce a deviation of 90° If the refractive index is greater than . (A) (B) (C) (D) None of these. Solution: (A) Hence a ray striking a glass-air surface from inside the prism will be reflected back if angle of incidence is not less than 45°. 90° Deviation is produced if the ray strikes on one of the sides normally as shown. Here, 1 = 45° and hence reflection takes place. 10. Convex lens of 10 cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination. (A) 15 cm (B) – 15 cm (C) 30 cm (D) – 30 cm Solution: (B) Here f1 = 10 cm, f2 = -6 cm, F = ? Use the formula 11. Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be (A) 5 cm (B) 7.5 cm (C) 10 cm (D) 2.5 cm Solution (A) x 2cm = 5 cm. 12. A double convex lens of focal length 20 cm is made of glass of refractive index 3/2. When placed completely in water , its focal length will be (A) 80 cm (B) 15 cm (C) 17.7 cm (D) 22.5 cm Solution (A) fw = 4 x fa = 4 x 20 = 80 cms. 13. Find the minimum angle of deviation for a prism with angle A = 60° and  = 1.5. (A) 37 (B) 40 (C) 55 (D) 65 Solution: (A) Minimum Deviation The angle of minimum deviation occurs when i = e and r1 = r2 and is given by: Substituting  = 1.5 and A = 60°, 14. Find the maximum angle of deviation for a prism with angle A = 60° and  = 1.5. (A) 50 (B) 58 (C) 64 (D) 60 Solution: (B) Maximum Deviation The deviation is maximum when i = 90° or e = 90° that is at grazing incidence or grazing emergence. Let i = 90° Sin e = u sin r2 = 1.5 sin 18° Deviation = = i + e – A = 90° + 28° – 60° = 58° CMP : Inside a substance such as glass or water, light travels more slowly than it does in a vacuum. If c denotes the speed of light in a vacuum and v denotes its speed through some other substance, then v = Where n is a constant called the index of refraction. To almost exact approximation, a substance’s index of refraction does not depend the wavelength of light. For instance, when red and blue light waves enter water, they both slow down by about the same amount. More precise measurements, however, reveal that n varies with wavelength. Table 1 presents some indices of refraction of Cutson glass, for different wavelengths of visible light. A nanometer (nm) is 10–9 meters. In a vacuum, light travesl at c = 3.0  108 m/s. Indices of refraction of Cutson glass (Table 1) approximate wavelength in vacuum (nm) n yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494 15. Inside Cutson glass (A) Orange light travels faster than yellow light (B) Yellow light travels faster than orange light (C) Orange and yellow light ravels equally fast (D) We cannot determine which colour of light travels faster Solution: (A) The velocity of light is inversely proportional to the index of refraction v = , So, the lower the n, the higher the v. According to table, Cutson glass has a slightly lower n for orange light than it has for yellow light. Therefore, inside Cutson glass, orange light travels slightly faster. 16. For blue green of wavelength 520 nm, the index of refraction of Cutson glass is probably closest to: (A) 1.49 (B) 1.50 (C) 1.51 (D) 1.52 Solution: (C) Table suggests a linear relationship between wavelength and index of refraction. Starting with yellow light, we can extrapolate with linear relationship to blue green light. As compared to yellow light waves, blue green light waves are 60 nm shorter (520 nm vs. 580 nm). By reading table from bottom to top, we see that each 20 nm decreases in wavelength corresponds to a 0.002 increase in n. So, since yellow and blue green differ in wavelength by 60 nm, the corresponding n’s should differ by 0.006. nblue green = nyellow + 0.006 = 1.500 + 0.006 = 1.506 which is closer to 1.51 than it is to 1.50. You can also address this question by extrapolating table. blue green 520 1.506 540 1.504 560 1.502 yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494 17. For visible light, which graph bes expresses the index of refraction of Cutson glass as a function of the frequency of the light? (A) (B) (C) (D) Solution: (A) As just discussed, index of refraction decreases linearly with wavelength Graph b show a linear relationship. But the problem wants the relationship between frequency and n, not between wavelength and n. Since higher wavelengths correspond to lower frequencies, and vica versa, n increasing with frequency. To clarify this reasoning, we should review the relationship between wavelength and frequency. For all waves, not just light waves, v = f, where v denotes velocity,  denotes wavelength, and f denotes frequency. In a vacuum, all light waves travel at speed c. So, frequency and wavelength are inversely proportional. f = Higher wavelengths correspond to lower frequencies, and vice versa. Therefore, since n decreases at higher wavelengths, it increases at higher frequencies. Again, extending table 1 can help us spot this relationship. Using f = we can find the frequency of different colors of light. For instance fyellow = = 5.17  1014 Hz fyellow–orange = = 5.0  1014 Hz 18. Which of the following phenomena happens because n varies with wavelength? (A) A lens focuses light (B) A prism brakes sunlight into different colours (C) Total internal reflection ensures that light travels down a fiber optic cable (D) Light rays entering a pond change direction at the pond’s surface Solution: (B) 19. The time required for light to pass through a glass slab of 2 mm thickness is (A) 10–8 s (B) 10–6 s (C) 10–11 s (D) None of these Solution (C) 20. A spherical surface of curvature R separates air (=1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to (A) 2R (B) 5R (C) 3R (D) 1.5R Solution (B) 21. If the tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, calculate the focal length of the objective (A) 100 cm (B) 10 cm (C) 20 cm (D) 25 cm Solution (A) L = f0 + fe = 105 cm and M = f0 = 100 cm and fe = 5 cm. 22. Time of exposure for a photographic print is 10 seconds, when a lamp of 50 cd is placed at 1 m from it. Then another lamp of luminous intensity I is used, and is kept at 2 m from it. If the time of exposure now is 20 s, the value of I (in cd) is (A) 25 (B) 100 (C) 200 (D) 20 Solution (B) 23. How does refractive index () of a material vary with respect to wavelength () ? A and B are constants (A)  = A + B2 (B) (C)  = A + B (D) Solution (B) Cauchy's formula is 24. Time required for making a print at a distance of 40 cm from a 60 watt lamp is 12.8 second. If the distance is decreased to 25 cm, then time required in making the similar print will be (A) 15 sec (B) 10 sec (C) 5 sec (D) remains some. Solution (C) If prints are similar, But I1 = I2 25. An achromatic convergent doublet has power of +2D. If power of convex lens is + 5D, then ratio of the dispersive powers of a convergent and divergent lenses will be (A) 3 : 5 (B) 2 : 5 (C) 2 : 3 (D) None of these Solution (A) P = +2D 26. If a convex lens having focal length f and refractive index 1.5 is immersed in water then new focal length will be (w = 4/3) (A) f (B) 1.5 f (C) (D) 4 f Solution (D) . . . (1) . . . (2) 27. Radii of curvature of a convex lens are 20 cm and 30 cm and its focal length if 24 cm. Refractive index of the lens will be (A) 1.33 (B) 1.5 (C) 1.71 (D) None of these Solution (B) 28. Both radii of curvature of a convex lens are 20 cm and refractive index of the material of the lens is 1.5. Rays parallel to the axis of the lens will converge at (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm Solution (B) or f=20 cm i.e., rays will converge at 20 cm from the optical centre of the lens. 29. Eye piece of an astronomical telescope has focal length of 5 cm. If angular magnification in normal adjustment is 10, then dustance between eye piece and objective should be (A) 15 cm (B) 35 cm (C) 55 cm (D) 75 cm Solution (C) In normal adjustment, L = f0 + fe and L = 5 cm + 50 cm = 55 cm. 30. Power of a convex lens is + 5D (g= 1.5). When this lens is immersed in a liquid of refractive index , it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be (A) (B) (C) (D) none of these Solution (A)