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# 2.A: Geometric Optics and Image Formation (Answers)

## Conceptual Questions

1. Virtual image cannot be projected on a screen. You cannot distinguish a real image from a virtual image simply by judging from the image perceived with your eye.

3. Yes, you can photograph a virtual image. For example, if you photograph your reflection from a plane mirror, you get a photograph of a virtual image. The camera focuses the light that enters its lens to form an image; whether the source of the light is a real object or a reflection from mirror (i.e., a virtual image) does not matter.

5. No, you can see the real image the same way you can see the virtual image. The retina of your eye effectively serves as a screen.

7. The mirror should be half your size and its top edge should be at the level of your eyes. The size does not depend on your distance from the mirror.

9. when the object is at infinity; see the mirror equation

11. Yes, negative magnification simply means that the image is upside down; this does not prevent the image from being larger than the object. For instance, for a concave mirror, if distance to the object is larger than one focal distance but smaller than two focal distances the image will be inverted and magnified.

15. The focal length of the lens is fixed, so the image distance changes as a function of object distance.

17. Yes, the focal length will change. The lens maker’s equation shows that the focal length depends on the index of refraction of the medium surrounding the lens. Because the index of refraction of water differs from that of air, the focal length of the lens will change when submerged in water.

19. A relaxed, normal-vision eye will focus parallel rays of light onto the retina.

21. A person with an internal lens will need glasses to read because their muscles cannot distort the lens as they do with biological lenses, so they cannot focus on near objects. To correct nearsightedness, the power of the intraocular lens must be less than that of the removed lens.

23. Microscopes create images of macroscopic size, so geometric optics applies.

25. The eyepiece would be moved slightly farther from the objective so that the image formed by the objective falls just beyond the focal length of the eyepiece.

## Problems

27.

29. It is in the focal point of the big mirror and at the center of curvature of the small mirror.

31. $$\displaystyle f=\frac{R}{2}⇒R=+1.60m$$

33. $$\displaystyle d_o=27.3cm$$

35. Step 1: Image formation by a mirror is involved.

Step 2: Draw the problem set up when possible.

Step 3: Use thin-lens equations to solve this problem.

Step 4: Find f.

Step 5: Given: $$\displaystyle m=1.50,d_o=0.120m$$.

Step 6: No ray tracing is needed.

Step 7: Using $$\displaystyle m=\frac{d_i}{d_o},d_i=−0.180m$$. Then, $$\displaystyle f=0.360m$$.

Step 8: The image is virtual because the image distance is negative. The focal length is positive, so the mirror is concave.

37. a. for a convex mirror $$\displaystyle d_i<0⇒m>0.m=+0.111$$;

b. $$\displaystyle d_i=−0.334cm$$ (behind the cornea);

c. $$\displaystyle f=−0.376cm$$, so that $$\displaystyle R=−0.752cm$$

39. $$\displaystyle m=\frac{h_i}{h_o}=−\frac{d_i}{d_o}=−\frac{−d_o}{d_o}=\frac{d_o}{d_o}=1⇒h_i=h_o$$

41. $$\displaystyle m=−11.0$$ $$\displaystyle A′=0.110m^2$$ $$\displaystyle I=6.82kW/m^2$$

43. $$\displaystyle x_{2m}=−x_{2m−1},(m=1,2,3,...),$$

$$\displaystyle x_{2m+1}=b−x_{2m},(m=0,1,2,...),$$ with $$\displaystyle x_0=a.$$

45. $$\displaystyle d_i=−55cm;m=+1.8$$

47. $$\displaystyle d_i=−41cm,m=1.4$$

49. proof

51. a. $$\displaystyle \frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}⇒d_i=3.43m$$;

b. $$\displaystyle m=−33.33$$, so that $$\displaystyle (2.40×10^{−2}m)(33.33)=80.0cm,$$ and

$$\displaystyle (3.60×10^{−2}m)(33.33)=1.20m⇒0.800m×1.20m$$ or $$\displaystyle 80.0cm×120cm$$

53. a. $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ $$\displaystyle d_i=5.08cm$$;

b. $$\displaystyle m=−1.695×10^{−2}$$, so the maximum height is $$\displaystyle \frac{0.036m}{1.695×10^{−2}}=2.12m⇒100%$$;

c. This seems quite reasonable, since at 3.00 m it is possible to get a full length picture of a person.

55. a. $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}⇒d_o=2.55m$$;

b. $$\displaystyle \frac{h_i}{h_o}=−\frac{d_i}{d_o}⇒h_o=1.00m$$

57. a. Using $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$, $$\displaystyle d_i=−56.67cm$$. Then we can determine the magnification, $$\displaystyle m=6.67$$.

b. $$\displaystyle d_i=−190cm$$ and $$\displaystyle m=+20.0$$;

c. The magnification m increases rapidly as you increase the object distance toward the focal length.

59. $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$

$$\displaystyle d_I=\frac{1}{(1/f)−(1/d_o)}$$

$$\displaystyle \frac{d_i}{d_o}=6.667×10^{−13}=\frac{h_i}{h_o}$$

$$\displaystyle h_i=−0.933mm$$

61. $$\displaystyle d_i=−6.7cm$$

$$\displaystyle h_i=4.0cm$$

63. 83 cm to the right of the converging lens, $$\displaystyle m=−2.3,h_i=6.9cm$$

65. $$\displaystyle P=52.0D$$

67. $$\displaystyle \frac{h_i}{h_o}=−\frac{d_i}{d_o}⇒h_i=−h_o(\frac{d_i}{d_o})=−(3.50mm)(\frac{2.00cm}{30.0cm})=−0.233mm$$

69. a. $$\displaystyle P=+62.5D$$;

b. $$\displaystyle \frac{h_i}{h_o}=−\frac{d_i}{d_o}⇒h_i=−0.250mm$$;

c. $$\displaystyle h_i=−0.0800mm$$

71. $$\displaystyle P=\frac{1}{d_o}+\frac{1}{d_i}⇒d_o=28.6cm$$

73. Originally, the close vision was 51.0 D. Therefore, $$\displaystyle P=\frac{1}{d_o}+\frac{1}{d_i}⇒d_o=1.00m$$

75. originally, $$\displaystyle P=70.0D$$; because the power for normal distant vision is 50.0 D, the power should be decreased by 20.0 D

77. $$\displaystyle P=\frac{1}{d_o}+\frac{1}{d_i}⇒d_o=0.333m$$

79. a. $$\displaystyle P=52.0D$$;

b. $$\displaystyle P′=56.16D$$ $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=P⇒d_o=16.2cm$$

81. We need $$\displaystyle d_i=−18.5cm$$ when $$\displaystyle d_o=∞$$, so $$\displaystyle P=−5.41D$$

83. Let $$\displaystyle x$$ = far point ⇒$$\displaystyle P=\frac{1}{−(x−0.0175m)}+\frac{1}{∞}⇒−xP+(0.0175m)P=1⇒x=26.8cm$$

85. $$\displaystyle M=6×$$

87. $$\displaystyle M=(\frac{25cm}{L})(1+\frac{L−ℓ}{f})$$ $$\displaystyle L−ℓ=d_o$$ $$\displaystyle d_o=13cm$$

89. $$\displaystyle M=2.5×$$

91. $$\displaystyle M=−2.1×$$

93. $$\displaystyle M=\frac{25cm}{f}$$ $$\displaystyle M_{max}=5$$

95. $$\displaystyle M^{young}_{max}=1+\frac{18cm}{f}⇒f=\frac{18cm}{M^{young}_{max}−1}$$

$$\displaystyle M^{old}_{max}=9.8×$$

97. a. $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}$$$$=\frac{1}{f}⇒d_i=4.65cm⇒m=−30.01$$;

b. $$\displaystyle M_{net}=−240$$

99. a. $$\displaystyle \frac{1}{d^{obj}_o}+$$$$\frac{1}{d^{obj}_i}$$$$=\frac{1}{f^{obj}}$$$$⇒d^{obj}_i=18.3cm$$ behind the objective lens;

b. $$\displaystyle m^{obj}=−60.0$$;

c. $$\displaystyle d^{eye}_o=1.70cm$$

$$\displaystyle d^{eye}_i=−11.3cm$$;

d. $$\displaystyle M^{eye}=13.5$$;

e. $$\displaystyle M_{net}=−810$$

101. $$\displaystyle M=−40.0$$

103. $$\displaystyle f^{obj}=\frac{R}{2},M=−1.67$$

105. $$\displaystyle M=−\frac{f^{obj}}{f^{eye}},f^{eye}=+10.0cm$$

109. 12 cm to the left of the mirror, $$\displaystyle m=3/5$$

111. 27 cm in front of the mirror, $$\displaystyle m=0.6,h_i=1.76cm$$, orientation upright

113. The following figure shows three successive images beginning with the image $$\displaystyle Q_1$$ in mirror $$\displaystyle M_1$$. $$\displaystyle Q_1$$ is the image in mirror $$\displaystyle M_1$$, whose image in mirror $$\displaystyle M_2$$ is $$\displaystyle Q_{12}$$ whose image in mirror $$\displaystyle M_1$$ is the real image $$\displaystyle Q_{121}$$.

115. 5.4 cm from the axis

117. Let the vertex of the concave mirror be the origin of the coordinate system. Image 1 is at −10/3 cm (−3.3 cm), image 2 is at −40/11 cm (−3.6 cm). These serve as objects for subsequent images, which are at −310/83 cm (−3.7 cm), −9340/2501 cm (−3.7 cm), −140,720/37,681 cm (−3.7 cm). All remaining images are at approximately −3.7 cm.

119.

121. Figure shows from left to right: an object with base O on the axis and tip P. A bi-concave lens with focal point F1 and F2 on the left and right respectively and a concave mirror with center of curvature C. Two rays originate from P and diverge through the bi-concave lens. Their back extensions converge between F1 and the lens to form image Q1. Two rays originating from the tip of Q1 strike the mirror, are reflected and converge at Q2 between C and the mirror.

123. −5 D

125. 11

127. a.

b.

c.

d. similar to the previous picture but with point P outside the focal length;

e. Repeat (a)–(d) for a point object off the axis. For a point object placed off axis in front of a concave mirror corresponding to parts (a) and (b), the case for convex mirror left as exercises.

129. $$\displaystyle d_i=−10/3cm,h_i=2cm$$, upright

131. proof

133.

Triangles BAO and $$\displaystyle B_1A_1O$$ are similar triangles. Thus, $$\displaystyle \frac{A_1B_1}{AB}=\frac{d_i}{d_o}$$. Triangles NOF and $$\displaystyle B_1A_1F$$ are similar triangles. Thus, $$\displaystyle \frac{NO}{f}=\frac{A_1B_1}{d_i−f}$$. Noting that $$\displaystyle NO=AB$$ gives $$\displaystyle \frac{AB}{f}=\frac{A_1B_1}{d_i−f}$$ or $$\displaystyle \frac{AB}{A_1B_1}=\frac{f}{d_i−f}$$. Inverting this gives $$\displaystyle \frac{A_1B_1}{AB}=\frac{d_i−f}{f}$$. Equating the two expressions for the ratio $$\displaystyle \frac{A_1B_1}{AB}$$ gives $$\displaystyle \frac{d_i}{d_o}=\frac{d_i−f}{f}$$. Dividing through by $$\displaystyle d_i$$ gives $$\displaystyle \frac{1}{d_o}=\frac{1}{f}−\frac{1}{d_i}$$ or $$\displaystyle \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$.

135. 70 cm

137. The plane mirror has an infinite focal point, so that $$\displaystyle d_i=−d_o$$. The total apparent distance of the man in the mirror will be his actual distance, plus the apparent image distance, or $$\displaystyle d_o+(−d_i)=2d_o$$. If this distance must be less than 20 cm, he should stand at $$\displaystyle d_o=10cm$$.

139. Here we want $$\displaystyle d_o=25cm−2.20cm=0.228m$$. If $$\displaystyle x=$$near point, $$\displaystyle d_i=−(x−0.0220m)$$. Thus, $$\displaystyle P=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{0.228m}+\frac{1}{x−0.0220m}$$. Using $$\displaystyle P=0.75D$$ gives $$\displaystyle x=0.253m$$, so the near point is 25.3 cm.

141. Assuming a lens at 2.00 cm from the boy’s eye, the image distance must be $$\displaystyle d_i=−(500cm−2.00cm)=−498cm$$. For an infinite-distance object, the required power is $$\displaystyle P=\frac{1}{d_i}=−0.200D$$. Therefore, the $$\displaystyle −4.00D$$ lens will correct the nearsightedness.

143. $$\displaystyle 87μm$$

145. Use, $$\displaystyle M_{net}=−\frac{d^{obj}_i(f^{eye}+25cm)}{f^{obj}f^{eye}}$$. The image distance for the objective is dobji=−Mnetfobjfeyefeye+25cmdiobj=−Mnetfobjfeyefeye+25cm. Using fobj=3.0cm,feye=10cm,andM=−10fobj=3.0cm,feye=10cm,andM=−10 gives $$\displaystyle d^{obj}_i=8.6cm$$. We want this image to be at the focal point of the eyepiece so that the eyepiece forms an image at infinity for comfortable viewing. Thus, the distance d between the lenses should be $$\displaystyle d=f^{eye}+d^{obj}_i=10cm+8.6cm=19cm$$

147. a. focal length of the corrective lens $$\displaystyle f_c=−80cm$$;

b. −1.25 D

149. $$\displaystyle 2×10^{16}km$$

151. $$\displaystyle 105m$$

### Contributors

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).