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
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
