Sign convention used throughout: Cartesian sign convention, distances measured from the pole/optical centre, positive in the direction of incident light. Mirror formula 1/v + 1/u = 1/f; lens formula 1/v − 1/u = 1/f.
Exercises
(9.1) A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer
Given: h = 2.5 cm, u = −27 cm, R = −36 cm
⇒ f = R/2 = −18 cm.
1/v = 1/f − 1/u
⇒ 1/v = −1/18 + 1/27 = (−3 + 2)/54 = −1/54
⇒ v = −54 cm
The screen must be placed 54 cm in front of the mirror.
m = −v/u
⇒ m = −(−54)/(−27) = −2
⇒ h′ = m × h = −2 × 2.5 = −5 cm
The image is real, inverted and magnified (5 cm tall). If the candle is moved closer to the mirror (but still beyond the focus), the image moves farther away, so the screen has to be moved away from the mirror. Once the candle comes inside the focus (closer than 18 cm), the image becomes virtual and cannot be caught on a screen at all.
(9.2) A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer
Given: h = 4.5 cm, u = −12 cm, f = +15 cm.
1/v = 1/f − 1/u
⇒ 1/v = 1/15 + 1/12 = (4 + 5)/60 = 9/60
⇒ v = 6.67 cm (behind the mirror)
m = −v/u
⇒ m = −6.67/(−12) ≈ 0.56
⇒ h′ = 0.56 × 4.5 ≈ 2.5 cm
The image is virtual, erect and diminished, 6.67 cm behind the mirror. As the needle is moved farther away, the image moves towards the focus (but never beyond it) and keeps getting smaller.
(9.3) A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer
n = real depth / apparent depth
⇒ n = 12.5/9.4
⇒ n ≈ 1.33
For the new liquid: apparent depth = 12.5/1.63 ≈ 7.67 cm.
The image of the needle is now raised by 9.4 − 7.67 = 1.73 cm, so the microscope must be moved up by about 1.73 cm to refocus.
(9.4) Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].
Answer
From Fig. (a), air to glass: angle in air 60°, angle in glass 35°
⇒ n_g = sin 60°/sin 35° = 0.866/0.574 ≈ 1.51
From Fig. (b), air to water: angle in air 60°, angle in water 47°
⇒ n_w = sin 60°/sin 47° = 0.866/0.731 ≈ 1.18
Fig. (c), water to glass: n_w sin 45° = n_g sin r
⇒ sin r = (1.18 × 0.707)/1.51 ≈ 0.553
⇒ r ≈ 33.7°
The angle of refraction in glass is about 34°.
(9.5) A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer
Light escapes only where the angle of incidence at the surface is less than the critical angle i_c, so it emerges through a circle of radius r = h tan i_c.
sin i_c = 1/n = 1/1.33 = 0.752
⇒ i_c ≈ 48.75°
⇒ tan i_c ≈ 1.14
r = 0.80 × 1.14 → r ≈ 0.91 m
Area = πr²
⇒ A = 3.14 × (0.91)²
⇒ A ≈ 2.6 m²
(9.6) A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer
Given: A = 60°, D_m = 40°.
n = sin[(A + D_m)/2] / sin(A/2)
⇒ n = sin 50°/sin 30° = 0.766/0.5
⇒ n ≈ 1.532
In water, the relative refractive index n′ = 1.532/1.33 ≈ 1.152.
sin[(60° + D′_m)/2] = n′ sin 30° = 1.152 × 0.5 = 0.576
⇒ (60° + D′_m)/2 ≈ 35.2°
⇒ D′_m ≈ 10.3°
(9.7) Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer
Given: n = 1.55, f = 20 cm, R₁ = R, R₂ = −R.
Lens maker's formula: 1/f = (n − 1)(1/R₁ − 1/R₂) = (n − 1)(2/R)
⇒ R = 2(n − 1)f
⇒ R = 2 × 0.55 × 20
⇒ R = 22 cm
(9.8) A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Answer
Point P acts as a virtual object on the far side of the lens: u = +12 cm.
(a) f = +20 cm: 1/v = 1/f + 1/u = 1/20 + 1/12 = (3 + 5)/60 = 8/60
⇒ v = 7.5 cm. The beam converges at a point 7.5 cm from the lens (on the side of P).
(b) f = −16 cm: 1/v = −1/16 + 1/12 = (−3 + 4)/48 = 1/48
⇒ v = 48 cm. The beam converges at a point 48 cm from the lens.
(9.9) An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Answer
Given: u = −14 cm, f = −21 cm, h = 3.0 cm.
1/v = 1/f + 1/u
⇒ 1/v = −1/21 − 1/14 = (−2 − 3)/42 = −5/42
⇒ v = −8.4 cm
m = v/u
⇒ m = −8.4/−14 = 0.6
⇒ h′ = 0.6 × 3.0 = 1.8 cm
The image is virtual, erect, diminished (1.8 cm), formed 8.4 cm from the lens on the same side as the object. As the object is moved farther away, the image moves towards the focus (21 cm) and becomes smaller, but it never goes beyond the focus.
(9.10) What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer
1/F = 1/f₁ + 1/f₂
⇒ 1/F = 1/30 − 1/20 = (2 − 3)/60 = −1/60
⇒ F = −60 cm
The combination behaves as a diverging (concave) lens of focal length 60 cm.
(9.11) A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Answer
Given: f_o = 2.0 cm, f_e = 6.25 cm, separation L = 15 cm.
(a) Eyepiece: v_e = −25 cm
⇒ 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/6.25 = −0.04 − 0.16 = −0.20
⇒ u_e = −5 cm
Image by objective: v_o = 15 − 5 = 10 cm.
Objective: 1/u_o = 1/v_o − 1/f_o = 1/10 − 1/2 = −0.4
⇒ u_o = −2.5 cm. The object is placed 2.5 cm from the objective.
m = (v_o/|u_o|)(1 + D/f_e)
⇒ m = (10/2.5)(1 + 25/6.25)
⇒ m = 4 × 5 = 20
(b) Final image at infinity: u_e = −f_e = −6.25 cm
⇒ v_o = 15 − 6.25 = 8.75 cm
1/u_o = 1/8.75 − 1/2 = 0.1143 − 0.5 = −0.3857
⇒ u_o ≈ −2.59 cm. The object is placed about 2.59 cm from the objective.
m = (v_o/|u_o|)(D/f_e)
⇒ m = (8.75/2.59)(25/6.25)
⇒ m ≈ 3.38 × 4 ≈ 13.5
(9.12) A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer
Given: f_o = 0.8 cm, u_o = −0.9 cm, f_e = 2.5 cm, D = 25 cm (final image at near point).
Objective: 1/v_o = 1/f_o + 1/u_o = 1/0.8 − 1/0.9 = 1.25 − 1.111 = 0.139
⇒ v_o = 7.2 cm
Eyepiece: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/2.5 = −0.44
⇒ u_e ≈ −2.27 cm
Separation = v_o + |u_e|
⇒ 7.2 + 2.27 ≈ 9.47 cm
m = (v_o/|u_o|)(1 + D/f_e)
⇒ m = (7.2/0.9)(1 + 25/2.5)
⇒ m = 8 × 11 = 88
(9.13) A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer
In normal adjustment: m = f_o/f_e
⇒ m = 144/6.0
⇒ m = 24
Separation L = f_o + f_e
⇒ L = 144 + 6
⇒ L = 150 cm
(9.14) (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.
Answer
(a) m = f_o/f_e
⇒ m = 1500 cm / 1.0 cm
⇒ m = 1500
(b) Angle subtended by the moon α = D_moon/r
⇒ α = 3.48 × 10⁶ / 3.8 × 10⁸ ≈ 9.16 × 10⁻³ rad
The objective forms the image in its focal plane, and the image subtends the same angle at the objective: d = α f_o
⇒ d = 9.16 × 10⁻³ × 15
⇒ d ≈ 0.137 m = 13.7 cm
(9.15) Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer
Mirror equation: 1/v = 1/f − 1/u. For a real object u is negative.
(a) Concave mirror: f < 0. Object between f and 2f means 2f < u < f (all negative), i.e. |f| < |u| < 2|f|.
1/v = −1/|f| + 1/|u|. Since |u| > |f|, 1/|u| < 1/|f|
⇒ 1/v < 0
⇒ v is negative, so the image is real (in front of the mirror).
Also |u| < 2|f|
⇒ 1/|u| > 1/(2|f|)
⇒ 1/|v| = 1/|f| − 1/|u| < 1/|f| − 1/(2|f|) = 1/(2|f|)
⇒ |v| > 2|f|. The real image lies beyond 2f.
(b) Convex mirror: f > 0 and u < 0
⇒ 1/v = 1/f + 1/|u| > 0
⇒ v is always positive. The image is always behind the mirror, i.e. virtual, wherever the object is.
(c) From (b), 1/v = 1/f + 1/|u| > 1/f
⇒ v < f: the image lies between the pole and the focus.
Also 1/v > 1/|u|
⇒ v < |u|
⇒ m = −v/u = v/|u| < 1: the image is always diminished.
(d) Concave mirror (f < 0) with the object between pole and focus: |u| < |f|.
1/v = −1/|f| + 1/|u|; since 1/|u| > 1/|f|, 1/v > 0
⇒ v is positive, so the image is virtual (behind the mirror).
1/v = 1/|u| − 1/|f| < 1/|u|
⇒ v > |u|
⇒ m = v/|u| > 1: the image is enlarged (and erect).
(9.16) A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Answer
Apparent shift = t(1 − 1/n)
⇒ shift = 15 × (1 − 1/1.5)
⇒ shift = 15 × 1/3 = 5 cm
The pin appears raised by 5 cm. The answer does not depend on the position of the slab (for small angles of viewing), since the shift depends only on the thickness and refractive index.
(9.17) (a) Figure 9.28 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?
Answer
(a) Critical angle at the fibre–covering boundary: sin i_c = 1.44/1.68 = 0.857
⇒ i_c ≈ 59°.
Total reflection needs the angle at the wall i′ ≥ 59°, so the refraction angle at the end face r = 90° − i′ must be ≤ 31°.
At the end face (air to fibre): sin i = 1.68 sin r ≤ 1.68 × sin 31° = 1.68 × 0.515 ≈ 0.865
⇒ i ≤ 60°.
Total internal reflection occurs for incident angles 0 < i < 60° with the axis.
(b) Without covering, the outside medium is air: sin i_c = 1/1.68
⇒ i_c ≈ 36.5°, so r can be up to 90° − 36.5° = 53.5°. Then 1.68 sin 53.5° ≈ 1.35, which exceeds 1, so even an incident angle of 90° gives r below 53.5°. All rays entering the end face (0 < i < 90°) suffer total internal reflection.
(9.18) The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Answer
For a real image, the object–image distance D must satisfy D ≥ 4f (the minimum D = 4f occurs when u = v = 2f).
f_max = D/4
⇒ f_max = 3/4
⇒ f_max = 0.75 m
(9.19) A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer
Given: D = 90 cm, d = 20 cm. In the two positions u and v interchange: |u| + v = 90 and v − |u| = 20
⇒ v = 55 cm, |u| = 35 cm.
1/f = 1/v − 1/u = 1/55 + 1/35 = (7 + 11)/385 = 18/385
⇒ f ≈ 21.4 cm
(Equivalently f = (D² − d²)/4D = (8100 − 400)/360 ≈ 21.4 cm.)
(9.20) (a) Determine the 'effective focal length' of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer
(a) f₁ = +30 cm (convex), f₂ = −20 cm (concave), d = 8 cm.
Parallel light incident on the convex lens: it would focus 30 cm beyond the convex lens, i.e. 22 cm beyond the concave lens
⇒ virtual object for concave lens u = +22 cm.
1/v = 1/f₂ + 1/u = −1/20 + 1/22 = (−11 + 10)/220 = −1/220
⇒ v = −220 cm. The emerging beam diverges from a point 220 cm to the left of the concave lens, i.e. 216 cm from the centre of the system.
Parallel light incident on the concave lens: it diverges from a point 20 cm in front of the concave lens, i.e. u = −(20 + 8) = −28 cm for the convex lens.
1/v = 1/f₁ + 1/u = 1/30 − 1/28 = (14 − 15)/420 = −1/420
⇒ v = −420 cm. The beam now appears to diverge from a point 420 cm from the convex lens, i.e. 416 cm from the centre.
The answer depends on which side the light enters, so the notion of a single effective focal length is not useful for this system.
(b) Convex lens: u = −40 cm
⇒ 1/v₁ = 1/30 − 1/40 = 1/120
⇒ v₁ = 120 cm
⇒ m₁ = v₁/u = 120/−40 = −3
Concave lens: this image is 120 − 8 = 112 cm beyond it, a virtual object u = +112 cm
⇒ 1/v₂ = −1/20 + 1/112 = (−28 + 5)/560 = −23/560
⇒ v₂ ≈ −24.3 cm
⇒ m₂ = v₂/u = −24.3/112 ≈ −0.217
Total magnification m = m₁ × m₂
⇒ m = (−3) × (−0.217)
⇒ m ≈ 0.652
Image size = 0.652 × 1.5
⇒ ≈ 0.98 cm (virtual, erect, 24.3 cm in front of the concave lens).
(9.21) At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Answer
At the second face the angle of incidence must equal the critical angle: sin r₂ = 1/1.524
⇒ r₂ ≈ 41.0°.
r₁ + r₂ = A → r₁ = 60° − 41.0° = 19.0°
At the first face: sin i = n sin r₁
⇒ sin i = 1.524 × sin 19.0° ≈ 1.524 × 0.325 ≈ 0.495
⇒ i ≈ 29.7°
For any angle of incidence less than about 29.7°, the ray is totally reflected at the second face.
(9.22) A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Answer
(a) u = −9 cm, f = +9 cm
⇒ 1/v = 1/f + 1/u = 1/9 − 1/9 = 0
⇒ v = ∞
The card is at the focus, so the virtual image is formed at infinity. The linear magnification m = v/u is infinitely large, and the area of each square in the image is likewise not finite. Linear magnification is not a meaningful measure here.
(b) With the image at infinity, each square of side h subtends an angle h/f = h/9 at the eye. Seen unaided at the near point it subtends h/25. Magnifying power = (h/9)/(h/25) = 25/9 ≈ 2.8
(c) No. The linear magnification (ratio of image size to object size) is infinite, while the magnifying power (ratio of angles subtended at the eye) is about 2.8. The two are equal only when the image is formed at the near point (25 cm).
(9.23) (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.
Answer
(a) Maximum magnifying power is obtained when the image is at the near point: v = −25 cm, f = 9 cm.
1/u = 1/v − 1/f = −1/25 − 1/9 = −34/225
⇒ u ≈ −6.6 cm. The lens should be held about 6.6 cm from the card.
(b) m = v/u → m = 25/6.62 → m ≈ 3.8
(c) Magnifying power = D/|u| = 25/6.62 ≈ 3.8 (equivalently 1 + D/f = 1 + 25/9 ≈ 3.8). Yes, they are equal here, because when the image is at the near point (25 cm), the angle subtended by the image at the eye (h′/25) divided by the angle the object would subtend at 25 cm (h/25) is just h′/h, the linear magnification.
(9.24) What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Answer
Area magnification = 6.25
⇒ linear magnification m = √6.25 = 2.5, so v = 2.5u.
1/v − 1/u = 1/f
⇒ 1/(2.5u) − 1/u = 1/9
⇒ −1.5/(2.5u) = 1/9
⇒ u = −5.4 cm, v = 2.5 × (−5.4) = −13.5 cm
The card should be 5.4 cm from the lens. The virtual image is then only 13.5 cm from the eye, closer than the near point (25 cm), so the squares cannot be seen distinctly with the eye very close to the magnifier.
(9.25) Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer
(a) Without the lens, the eye cannot see an object clearly if it is closer than 25 cm, so the largest angle it can subtend while being seen distinctly is the angle at 25 cm. The magnifier lets us bring the object much closer to the eye (inside 25 cm) while its virtual image is formed at or beyond 25 cm, where it can be seen clearly. The object therefore subtends a larger angle than it could when viewed distinctly without the lens; that is the angular magnification.
(b) Yes. If the eye moves back from the lens, the angle subtended at the eye by the image becomes slightly smaller than the angle subtended at the lens, so the angular magnification decreases a little. The field of view also shrinks.
(c) Lenses of very short focal length are strongly curved and hard to grind accurately. More importantly, spherical and chromatic aberrations grow sharply as the focal length decreases, so the image becomes blurred and coloured. In practice a simple convex lens cannot give useful magnifying power much beyond about 10.
(d) The magnifying power is m ≈ (L/f_o)(D/f_e). The objective magnification L/f_o is large when f_o is small, and the eyepiece magnification D/f_e is large when f_e is small. A short objective focal length also lets the object be placed very close to the objective, giving a large magnified intermediate image.
(e) The eyepiece forms an image of the objective, called the eye-ring (exit pupil). All the light from the object that passes through the objective also passes through this ring. Placing the eye at the eye-ring collects the maximum light and gives the widest field of view. Its exact position depends on the separation of the lenses and on f_e; in practice it is a short distance just beyond the eyepiece focal point (about a centimetre or so for typical eyepieces), and eyecups are designed to hold the eye there.
(9.26) An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Answer
Take the final image at the near point (25 cm).
Eyepiece magnification m_e = 1 + D/f_e
⇒ m_e = 1 + 25/5 = 6. So the objective must give m_o = 30/6 = 5, i.e. v_o = 5|u_o|.
Objective: 1/v_o − 1/u_o = 1/f_o
⇒ 1/(5x) + 1/x = 1/1.25 (with |u_o| = x)
⇒ 6/(5x) = 0.8
⇒ x = 1.5 cm
⇒ v_o = 7.5 cm
Eyepiece: 1/u_e = 1/v_e − 1/f_e = −1/25 − 1/5 = −6/25
⇒ u_e ≈ −4.17 cm
Separation = v_o + |u_e|
⇒ 7.5 + 4.17 ≈ 11.7 cm
So place the object 1.5 cm in front of the objective and keep the objective and eyepiece about 11.7 cm apart.
(9.27) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)?
Answer
(a) m = f_o/f_e → m = 140/5 → m = 28
(b) m = (f_o/f_e)(1 + f_e/D)
⇒ m = 28 × (1 + 5/25)
⇒ m = 28 × 1.2 = 33.6
(9.28) (a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25 cm?
Answer
(a) In normal adjustment L = f_o + f_e
⇒ L = 140 + 5
⇒ L = 145 cm
(b) Angle subtended by the tower α = 100/3000 = 1/30 rad. Image by objective h = α f_o
⇒ h = 140/30
⇒ h ≈ 4.7 cm
(c) For the final image at 25 cm, the eyepiece magnification m_e = 1 + D/f_e = 1 + 25/5 = 6
⇒ final height = 6 × 4.67 ≈ 28 cm
(9.29) A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Answer
Large (concave, objective) mirror: f₁ = 220/2 = 110 mm. Small (convex, secondary) mirror: f₂ = 140/2 = 70 mm. Separation = 20 mm.
Parallel rays reflected by the large mirror would converge at its focus, 110 mm from it. The small mirror intercepts them 20 mm from the large mirror, so the point of convergence lies 110 − 20 = 90 mm behind the small mirror: a virtual object, u = +90 mm (measured along the direction of the incident light on the small mirror).
For the convex mirror f = +70 mm: 1/v = 1/f − 1/u = 1/70 − 1/90 = (9 − 7)/630 = 2/630
⇒ v = +315 mm
The final image is formed 315 mm from the small mirror, on the side of the small mirror away from the large mirror.
(9.30) Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Answer
When a mirror turns through θ, the reflected ray turns through 2θ = 7°.
d = D tan 2θ
⇒ d = 1.5 × tan 7°
⇒ d = 1.5 × 0.1228
⇒ d ≈ 0.184 m = 18.4 cm
(9.31) Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Answer
The image coincides with the needle when light leaves the lens system parallel, strikes the mirror normally and retraces its path, i.e. when the needle is at the focus of the lens system.
Without liquid: f₁ = 30 cm. With liquid: focal length of combination F = 45 cm.
Glass lens (equiconvex, R₁ = R, R₂ = −R): 1/f₁ = (1.5 − 1)(2/R)
⇒ 1/30 = 1/R
⇒ R = 30 cm
Liquid lens: 1/F = 1/f₁ + 1/f₂
⇒ 1/f₂ = 1/45 − 1/30 = (2 − 3)/90
⇒ f₂ = −90 cm
The liquid forms a plano-concave lens with one surface of radius 30 cm and the other plane: 1/f₂ = (μ − 1)(−1/R)
⇒ −1/90 = −(μ − 1)/30
⇒ μ − 1 = 1/3
⇒ μ ≈ 1.33
Important Extra Questions
State the conditions for total internal reflection.
Answer
(i) Light must travel from an optically denser medium towards a rarer medium. (ii) The angle of incidence in the denser medium must be greater than the critical angle for the pair of media, where sin i_c = n_rarer/n_denser.
Write the relation between the focal length and the radius of curvature of a spherical mirror.
Answer
f = R/2, valid for paraxial rays (rays close to the principal axis).
What is the power of a lens? Find the power of a concave lens of focal length 25 cm.
Answer
Power P = 1/f (f in metres); its unit is the dioptre (D). For f = −0.25 m, P = 1/(−0.25) = −4 D.
Why are the objectives of large astronomical telescopes usually mirrors rather than lenses?
Answer
A mirror objective has no chromatic aberration, a paraboloidal mirror removes spherical aberration, a large mirror is lighter and easier to support (from its back) than a large lens, and only one surface has to be ground and polished.
Why does white light split into colours when it passes through a prism?
Answer
The refractive index of glass depends on wavelength: it is greater for violet than for red. So different colours are deviated by different amounts, violet the most and red the least. This splitting is called dispersion.