Chapter #5 Solutions - Optics - Ajoy Ghatak - 1st Edition

1. Consider a system of two thin convex lenses of focal lengths 10 cm and 30 cm separated by a distance of 20 cm in air.(a) Determine the system matrix elements and the positions of the unit planes(b) Assume a parallel beam of light incident from the left. Use Eq. (67) and the positions of the unit planes to determine the image point. Using the unit planes draw the ray diagram.[Ans. (a) a =1/15, b =1/3, c = -1, d = -20; the first convex lens is in the middle of the two unit planes. (b) The final image is virtual and is 15 cm away (on the left) from the second lens.] Get solution

2. Consider a thick biconvex lens whose magnitudes of the radii of curvature of the first and second surfaces are 45 cm and 30 cm respectively. The thickness of the lens is 5 cm and the refractive index of the material of the lens is 1.5. Determine the elements of system matrix and positions of the unit planes and use Eq. (67) to determine the image point of an object at a distance of 90 cm from the first surface.[ Ans. a = 0.02716, b = 0.9444, c = 0.9630, d = –3.3333, du1 =2.0455, du2 = –1.3636. Final image at a distance of 60 cm from the second surface.] Get solution

3. Consider a hemisphere of radius 20 cm and refractive index 1.5. If H1 and H2 denote the positions of the first and second principal points, then show that AH1 = 13.3 cm and that H2 lies on the second surface as shown in Fig. 4.13. Further, show that the focal length is 40 cm.... Get solution

4. Consider a thick lens of the form shown in Fig. 4.14; the radii of curvature of the first and second surfaces are – 10 cm and +20 cm respectively and the thickness of the lens is 1.0 cm. The refractive index of the material of the lens is 1.5. Determine the positions of the principal planes.[ Ans. du1= 20/91 cm, du2 = – 40/91 cm]... Get solution

5. Consider a combinations of two thin lenses of focal lengths f1 and f2 separated by a distance (f1 + f2). Show that the angular magnification of the lens combinations (which is just ...) is given by – f1/f2. Interpret the negative sign in the expression for magnification. Get solution

6. Consider a spherical refracting surface as shown in Fig. 3.12. Using matrix method show that for an object at a distance of ...from the surface, the image is virtual and at a distance of ...from the surface. Get solution

Chapter #4 Solutions - Optics - Ajoy Ghatak - 1st Edition

1. (a) Consider a thin biconvex lens (as shown in Fig. 3.18) made of a material whose refractive index is 1.5. The radii of curvature of the first and second surfaces (R1 and R2) are +100 and –60 cm respectively. The lens is placed in air (i.e. n1 = n3 = 1). For an object at a distance of 100 cm from the lens, determine the position and linear magnification of the (paraxial) image. Also calculate x1 and x2 and verify Newton’s formula [Eq. (20)].(b) Repeat the calculations of the above problem when the object is at a distance of 50 cm.... Get solution

2. Consider a thin lens (made of a material of refractive index n2) having different media on the two sides; let n1 and n3 be the refractive indices of the media on the left and on the right of the lens respectively. Using Eq. (5) and considering successive refractions at the two surfaces, derive Eq. (14). Get solution

3. Referring again to Fig. 4.18 assume a biconvex lens with |R1| =100 cm, |R2| = 60 cm with n1 =1.0 but n3 =1.6. For u = –50 cm determine the position of the (paraxial) image. Also determine the first and second principal foci and verify Newton’s formula. Draw the ray diagram. Get solution

4. (a) In Fig 4.18, assume the convex lens to be replaced by a (thin) biconcave lens with |R1| =100 cm, |R2| = 60 cm. Assume n1 = n3 = 1 and n2 = 1.5. Determine the position of the image and draw an approximate ray diagram for u = – 100 cm.(b) In (a), assume n1 = n3 = 1.5 and n2 = 1.3. Repeat the calculations and draw the ray diagram. What is the qualitative difference between the systems in (a) and (b). Get solution

5. Consider an object of height 1 cm placed at a distance of 24 cm from a convex lens of focal length 15 cm (see Fig. 4.19). A concave lens of focal length – 20 cm is placed beyond the convex lens at a distance of 25 cm. Draw the ray diagram and determine the position and size of the final image....Fig. 4.19 An optical system consisting of a thin convex and a thin concave lens. All distances are measured in centimeters. Get solution

6. Consider a thick biconvex lens whose magnitude of the radii of curvature of the first and second surfaces are 45 and 30 cm respectively. The thickness of the lens is 5 cm and the refractive index of the material, it is made of, is 1.5. for an object of height 1 cm at a distance of 90 cm from the first surface, determine the position and size of the image. Draw the ray diagram for the axial point of the object. Get solution

7. In the above problem assume that the second surface is silvered so that it acts like a concave mirror. For an object of height 1 cm at a distance of 90 cm from the first surface determine the position and size of the image and draw the ray diagram. Get solution

8. Consider a sphere of radius 20 cm of refractive index 1.6 (see Fig. 4.20). Show that the paraxial focal point is at a distance of 6.7 cm from the point P2.... Get solution

9. Consider a hemisphere of radius 20 cm and refractive index 1.5. Show that parallel rays will focus at a point 40 cm from P2 (see Fig. 4.21).... Get solution

10. Consider a lens of thickness 1 cm made of a material of refractive index 1.5, placed in air. The radii of curvature of the first and second surfaces are +4 cm and –4 cm respectively. Determine the point at which parallel rays will focus. Get solution

Chapter #3 Solutions - Optics - Ajoy Ghatak - 1st Edition

1. In this and the following two problems we will use Fermat’s principle to derive laws governing paraxial image formation by spherical mirrors.Consider an object point O in front of a concave mirror whose center of curvature is at the point C. Consider an arbitrary point Q on the axis of the system and using a method similar to that used in Example 3.3, show that the optical path length Lop(= OS + SQ) is approximately given by...  (90)where the distances x, y and r and the angle θ are defined in Fig. 3.32; θ is assumed to be small. Determine the paraxial image point and show that the result is consistent with the mirror equation...   (91)where u and v are the object and image distance and R is the radius of curvature with the sign convention that all distances to the right of P are positive and to its left negative. Get solution

2. Fermat’s principle can also be used to determine the paraxial image points when the object forms a virtual image. Consider an object point O in front of the convex mirror SPM (see Fig. 3.33). One should now assume the optical path length Lop to be OS – SQ; the minus sign occurs because the rays at S point away from Q [see Example 3.4]. Show that...     (92)where the distances x, y and r and the angle θ are defined in Fig. 3.33. Show that the paraxial image is formed at y = y0 which is given by...     (93)which is consistent with Eq.(91) because whereas the object distance u is positive, the image distance v and the radius of curvature R are negative since the image point and the center of curvature lie on the left of the point P. Get solution

3. Proceeding as in the previous problem, use Fermat’s principle to determine the mirror equation for an object point at a distance less than R/2 from a concave mirror of radius of curvature R. Get solution

4. We next consider a point object O in front of a concave refracting surface SPM separating two media of refracting indices n1 and n2 [see Fig. 3.34]; C represents the center of curvature. In this case also one obtains a virtual image. Let Q represent an arbitrary point on the axis. We now have to consider the optical path length Lop = n1 OS – n2 SQ; show that it is given by...    (94)Also show that the above expression leads to the paraxial image point which is consistent with Eq.(10); we may note that u,v and R are all negative quantities because they are on the left of the refracting surface. Get solution

5. If we rotate an ellipse about its major axis we obtain what is known as an ellipsoid of revolution. Show by using Fermat’s principle that all rays parallel to the major axis of the ellipse will focus to one of the focal points of the ellipse (see Fig. 3.35), provided the eccentricity of the ellipse equals n1/n2.(Hint: Start with the condition that n2 AC′ = n1 QB + n2BC and show that the point B (whose coordinates are x and y) lies on the periphery of an ellipse). Get solution

6. C is the center of the reflecting sphere of radius R (see Fig. 3.36). P1 and P2 are twopoints on a diameter equidistant from the center. Obtain (a) the optical path length P1O+P2O as a function of θ and (b) find the values of θ for which P1OP2 is a ray path from reflection at the sphere. Get solution

7. SPM is a spherical refracting surface separating two media of refractive indices n1 and n2. (see Fig. 3.37). Consider an object point O forming a virtual image at the point I. We assume that all rays emanating from O appear to emanate from I so as to form a perfect image. Thus according to Fermat’s principle, we must haven1 OS– n2 SI= n1 OP– n2 PIwhere S is an arbitrary point on the refracting surface. Assuming the right hand side to be zero, show that the refracting surface is spherical, with the radius given by...    (95)Thus show that...    (96)where d1 and d2 are defined in Fig. 3.37; (see also sec. 4.10).[Hint: We consider a point C which is at a distance d1 from the point O and d2 from the point I. Assume the origin to be at O and let (x, y, z) represent the coordinates of the point S. Thus n1 (x2 + y2 + z2) ½ – n2 (x2 + y2 + Δ2) ½ = n1 (r + d1) – n2 (r + d2) = 0where Δ = d2 – d1. The above equation would give the equation of a sphere whose center is at a distance of n2r/n1 (= d1) from O.] Get solution

8. Referring to Fig. 3.38, if I represents a perfect image of the point O, show that the equation of the refracting surface (separating two media of refractive indices n1 and n2) is given by n1 [x2 + y2 + z2] ½ + n2 [x2 + y2 + (z2 – z)2] ½ = n1 z1 – n2 (z2 – z1) (97)where the origin is assumed to be at the point O and the coordinates of P and I are assumed to be (0,0, z1) and (0,0, z2) respectively. The surface corresponding to Eq.(97) is known as a Cartesian oval. Get solution

9. For the refractive index variation given by Eqs. (21) and (22), a ray is launched at x = .43m making an angle - π/60 with the z-axis (see Fig. 3.12). Calculate the value of x at which it will become horizontal. Get solution

10. For the refractive index variation given by Eqs. (21) and (22), a ray is launched at x = 2.8m such that it becomes horizontal at x = 0.2m (see Fig. 3.15). Calculate the angle that the ray will make with the z-axis at the launching point. Get solution

11. Consider a parabolic index medium characterized the following refractive index variation:  ...Assume n1 = 1.50, n2 = 1.48, ... = 50 μm. Calculate the value of Δ.(a) Assume rays launched on the axis at z = 0 (i.e., x = 0 when z = 0) with  ... = 1.495, 1.490, 1.485, 1.480, 1.475 and 1.470In each case calculate the angle that the ray initially makes with the z-axis (θ1) and plot the ray paths. In each case find the height at which the ray becomes horizontal.(b) Assume rays incident normally on the plane z = 0 at x = 0, ±10 μm, ±20 μm, ±30 μm, ±40 μm. Find the corresponding values of ..., calculate the focal length for each ray and qualitatively plot the ray paths. Get solution

12. In an inhomogeneous medium the refractive index is given by...Write down the equation of a ray (in the x-z plane) passing through the point (0, 0, 0) where its orientation with respect to x axis is 45°. Get solution

13. For the refractive index profile given by Eq. (23), show that Eq. (27) can be written in the form...  (98)where... and ...  (99)Integrate Eq. (98) to determine the ray paths. Get solution

14. Consider a graded index medium characterized by the following refractive indexdistribution...      (100)Substitute in Eq.(32) and integrate to obtain...    (101)Notice that the periodic length...is independent of the launching angle (see Fig. 3.32) and all rays rigorously take the same amount of time in propagating through a distance zp in the z – direction.[Hint: While carrying out the integration, make the substitution......Fig 3.40 Get solution

15. For z n =1For z > 0; ......n1 = 2.0; α = 15/16; a = 30 μmA ray is incident at the point A (x = x0 = 14 μm, z = 0) as shown in the Fig 3.40. (a)Calculate ...for the ray inside the graded index medium, (b) Calculate the maximum height h of the ray, (c) Calculate the angle θ that the ray makes with the z- axis at C, (d) Derive the equation of the ray path, (e) Calculate the time taken for the ray to traverse from B to C....Fig 3.41 Get solution

16. ...  ... where a = 2 mm(a) A ray is launched at 45° as shown in the figure.(b) Determine the ray path.(c) What is the time taken by the ray from A to B?[Ans. (a) ...; (b) ...] Get solution

Chapter #2 Solutions - Optics - Ajoy Ghatak - 1st Edition

1. An electron of energy 200 eV is passed through a circular hole of radius 10-4 cm. What is the uncertainty introduced in the momentum and also in the angle of emergence?[Ans: ...g cm/s; ... radians] Get solution

2. In continuation of the previous problem, what would be the corresponding uncertainty for a 0.1g lead ball thrown with a velocity 103 cm/sec through a hole 1 cm in radius?[Ans: ... radians] Get solution

3. A photon of wavelength 6000 Å is passed through a slit of width 0.2 mm(a) Calculate the uncertainty introduced in the angle of emergence.(b) The first minimum in the single slit diffraction pattern occurs at sin-1 (λ/b) where b is the width of the slit. Calculate this angle and compare with the angle obtained in part (a).[Ans: ... radians] Get solution

4. A 50 W bulb radiates light of wavelength 0.6 μm. Calculate the number of photons emitted per second.[Ans: ≈ 1.5 × 1020 photons/s] Get solution

5. Calculate the uncertainty in the momentum of a proton which is confined to a nucleus of radius equal to 10-13 cm. From this result, estimate the kinetic energy of the proton inside the nucleus. What would be the kinetic energy for an electron if it had to be confined within a similar nucleus? Get solution

6. The lifetime of the 2P state of the hydrogen atom is about 1.6 × 10-9 s. Use the time energy uncertainty relation to calculate the frequency width Δν.[Ans: ≈ 6 × 108 s-1] Get solution

7. A 1 W laser beam (of diameter 2 cm) falls normally on two circular holes each of diameter 0.05 cm as shown in Fig. 1.13. Calculate the average number of photons that will be found between the planes AB and PP′. Assume λ = 6 × 10–5 cm and the distance between the planes AB and PP′ to be 30 cm.[Ans. ≈ 4× 106 photons]... Get solution