31 worked answers · short steps first, explanations when you need them
Chapter 9 Exercises
Textbook pages 249–253
Question 1
A 2.5 cm candle is placed 27 cm in front of a concave mirror with radius of curvature 36 cm. Where should a screen be placed to get a sharp image? State the image's nature and size. What happens if the candle is moved closer to the mirror?
Steps only
f = −18 cm; 1/v = 1/f − 1/u = −1/18 + 1/27 = −1/54, so v = −54 cm.
m = −v/u = −2, giving an inverted image of height 5 cm.
Answer
Screen: 54 cm in front. Image: real, inverted, 5 cm high.
Question 2
A needle 4.5 cm long is placed 12 cm in front of a convex mirror with focal length 15 cm. Find the image position, magnification, and image length. What changes when the needle is moved farther away?
Steps only
1/v = 1/15 + 1/12 = 3/20, hence v = +6.67 cm.
m = −v/u = 0.556; image length = 0.556 × 4.5 = 2.5 cm.
Answer
Virtual upright image, 6.67 cm behind the mirror, 2.5 cm long; it shrinks as the needle moves away.
Question 3
A tank is 12.5 cm deep. A needle at its bottom appears 9.4 cm below the water surface. Find the refractive index of water. If water is replaced by a liquid of refractive index 1.63 at the same level, how far must a microscope be moved to focus on the needle?
Steps only
n = 12.5/9.4 ≈ 1.33; new apparent depth = 12.5/1.63 ≈ 7.67 cm.
9.4 − 7.67 = 1.73 cm upward shift.
Answer
n_water ≈ 1.33; move the microscope upward by 1.73 cm.
Question 4
Figures 9.27(a) and (b) show a ray incident from air at 60° on glass–air and water–air interfaces. Figure (c) shows a ray in water incident at 45° on a water–glass interface. Predict the angle of refraction in glass.
Steps only
n_glass/n_water = sin47°/sin35° ≈ 1.28. Then sin r = sin45°/1.28 ≈ 0.55, so r ≈ 33°.
Answer
About 33° to the normal.
Question 5
A point bulb is at the bottom of a 80 cm deep water tank. Water has refractive index 1.33. Find the area of the water surface through which light from the bulb can emerge.
Steps only
tan c = 1/√(1.33²−1) ≈ 1.14; r = 0.80×1.14 ≈ 0.91 m.
A = πr² ≈ 2.61 m².
Answer
About 2.61 m².
Question 6
A prism has angle 60° and minimum deviation 40°. Find its refractive index. What is the new minimum deviation if it is immersed in water of refractive index 1.33?
Steps only
n = sin50°/sin30° ≈ 1.532; n_rel = 1.532/1.33 = 1.152.
δ = 2sin⁻¹(1.152×sin30°)−60° ≈ 10.3°.
Answer
n ≈ 1.532; minimum deviation in water ≈ 10.3°.
Question 7
An equiconvex glass lens has refractive index 1.55 and focal length 20 cm in air. Find the radius of curvature of each face.
Steps only
1/f = 2(n−1)/R, so R = 2×0.55×20 = 22 cm.
Answer
22 cm for each face.
Question 8
A converging beam would meet at point P, 12 cm beyond a thin lens. Find where it will meet after the lens is inserted if the lens is (a) convex with focal length 20 cm, or (b) concave with focal length 16 cm.
Steps only
Use u=+12 cm. Convex: 1/v=1/20+1/12, so v=7.5 cm. Concave: 1/v=−1/16+1/12, so v=48 cm.
Answer
(a) 7.5 cm beyond; (b) 48 cm beyond.
Question 9
A 3 cm object is 14 cm in front of a concave lens of focal length 21 cm. Describe its image. What happens when the object is moved farther away?
Steps only
1/v=−1/21−1/14=−5/42, hence v=−8.4 cm. m=v/u=0.60 and image height=1.8 cm.
Answer
Virtual, upright image: 1.8 cm high, 8.4 cm from the lens on the object side.
Question 10
A convex lens of focal length 30 cm is in contact with a concave lens of focal length 20 cm. Find the effective focal length and say whether the pair converges or diverges light.
Steps only
P=1/0.30−1/0.20=−1.67 D, so F=−0.60 m=−60 cm.
Answer
−60 cm; diverging.
Question 11
A compound microscope has objective focal length 2 cm, eyepiece focal length 6.25 cm, and lens separation 15 cm. Find the object distance and magnifying power when the final image is (a) at the near point 25 cm, and (b) at infinity.
Steps only
At near point: uₑ=−5 cm, intermediate image 10 cm from objective, so uₒ=−2.5 cm. M=4×5=20.
At infinity: intermediate image 8.75 cm from objective, uₒ≈−2.59 cm. M≈3.375×4=13.5.
Answer
(a) Object 2.5 cm before objective; 20×. (b) Object 2.59 cm before objective; 13.5×.
Question 12
A person with a normal near point of 25 cm uses a compound microscope with objective focal length 8 mm and eyepiece focal length 2.5 cm. An object 9 mm from the objective is sharply focused. Find the lens separation and magnifying power when the final image is at the near point.
Steps only
Objective forms an image 7.2 cm away. Eyepiece object distance for a near-point image is 2.27 cm, so separation ≈9.47 cm.
M=(7.2/0.9)(1+25/2.5)=8×11=88.
Answer
Separation ≈9.47 cm; magnifying power ≈88×.
Question 13
A telescope has objective focal length 144 cm and eyepiece focal length 6 cm. Find its magnifying power and the distance between the lenses for normal adjustment (final image at infinity).
Steps only
M=144/6=24; L=144+6=150 cm.
Answer
24×; separation 150 cm.
Question 14
A large refracting telescope has objective focal length 15 m and eyepiece focal length 1 cm. Find its angular magnification. Also find the diameter of the Moon's image when the Moon is 3.48×10⁶ m across and 3.8×10⁸ m away.
Steps only
M=15/0.01=1500. Image diameter=15×(3.48×10⁶/3.8×10⁸)=0.137 m.
Answer
1500×; image diameter ≈13.7 cm.
Question 15
Use the mirror equation to show: (a) an object between F and 2F of a concave mirror has a real image beyond 2F; (b) a convex mirror always gives a virtual image; (c) a convex mirror always gives a diminished image between F and the pole; (d) an object between the pole and F of a concave mirror has a virtual, enlarged image.
Steps only
Concave: v=−Fx/(x−F) for x>F (real); if F<x<2F, |v|>2F. If x<F, v=Fx/(F−x)>0 and m=F/(F−x)>1.
Convex: v=Fx/(x+F)>0 and m=F/(x+F)<1, so its image is virtual, diminished, and between focus and pole.
Answer
The four stated image properties follow directly from the mirror equation.
Question 16
A pin is viewed vertically from above through a 15 cm thick glass slab of refractive index 1.5 placed on a table. The pin is 50 cm below the observer. How far does it appear raised?
Steps only
Apparent depth=15/1.5=10 cm; shift=15−10=5 cm.
Answer
5 cm upward.
Question 17
An optical fibre has core refractive index 1.68 and cladding refractive index 1.44. Find the largest angle an entering ray may make with the fibre axis and still be guided. What if the fibre has no cladding and is surrounded by air?
Steps only
sin θₘₐₓ=√(1.68²−1.44²)=0.865, so θₘₐₓ≈59.9°.
Without cladding, √(1.68²−1²)>1, so the air acceptance angle is limited to 90°.
Answer
About 60° with cladding; up to 90° without it (ideal case).
Question 18
Two opposite walls of a room are 3 m apart. What is the greatest possible focal length of a large convex lens that can form an image of an object on one wall on a screen at the other wall?
Steps only
For fixed u+v=3 m, uv is largest at u=v=1.5 m. Then f=uv/(u+v)=0.75 m.
Answer
0.75 m.
Question 19
An object and screen are 90 cm apart. A convex lens forms sharp images in two positions 20 cm apart. Find the lens focal length.
Steps only
f=(90²−20²)/(4×90)=21.39 cm.
Answer
About 21.4 cm.
Question 20
A convex lens of focal length 30 cm and a concave lens of focal length 20 cm are separated by 8 cm. (a) Find the effective focal length and discuss whether the direction of incidence matters. (b) A 1.5 cm object is 40 cm in front of the convex lens. Find the final image magnification and size.
Steps only
P=1/30−1/20−8/(30×−20)=−1/300 cm⁻¹, so F=−300 cm.
First lens gives v₁=120 cm. The second lens sees u₂=+112 cm and gives v₂≈−24.35 cm. Total m≈(−3)(−0.217)=+0.65; height≈0.98 cm.
Answer
F=−300 cm; final image is upright, virtual, and about 0.98 cm tall.
Question 21
A ray enters a prism of angle 60° and refractive index 1.524. It just undergoes total internal reflection at the second face. Find the angle of incidence at the first face.
Steps only
c=sin⁻¹(1/1.524)≈41°. Then r₁=60°−41°=19° and i=sin⁻¹(1.524 sin19°)≈29.7°.
Answer
About 30°.
Question 22
A 1 mm² square card is viewed through a magnifier of focal length 9 cm held close to the eye. The card is 9 cm from the lens. Find (a) the linear magnification and virtual-image area, (b) the angular magnification, and (c) whether these magnifications are equal.
Steps only
u=−f, so v=∞. There is no finite linear image area. Angular magnification=25/9≈2.78.
Answer
Image at infinity; linear size is unbounded. Angular magnification ≈2.78; the two are not equal.
Question 23
For the card and magnifier in Exercise 9.22, where should the lens be placed for maximum magnification while the image remains distinct? Find the magnification. Does it equal angular magnification?
Steps only
Set v=−25 cm: u≈−6.62 cm. Then |m|=25/6.62≈3.78=1+25/9.
Answer
Lens–card distance ≈6.62 cm; both magnifications ≈3.78×.
Question 24
Using the 1 mm² square card and 9 cm focal-length magnifier of Exercise 9.22, a virtual image has area 6.25 mm². Can a person with a normal near point see the squares distinctly if the eye is very close to the magnifier?
Steps only
√(6.25/1)=2.5 gives |m|=2.5. With f=9 cm, u=−5.4 cm and v=−13.5 cm.
Answer
No. The virtual image is inside the 25 cm near point.
Question 25
Answer these questions about a magnifying glass: (a) Why can a virtual image have a magnifying power even when its angle is the same as the object's angle? (b) What happens if the eye is moved farther from the lens? (c) Why can its focal length not be made arbitrarily small? (d) Why do a microscope's objective and eyepiece both have short focal lengths? (e) Where should the eye be placed relative to the eyepiece, and why?
Steps only
(a) It compares angles at a standard viewing distance. (b) Moving back reduces field and brightness. (c) Very short f causes distortion and little working room. (d) Short f in both lenses increases total magnification. (e) Look through the exit pupil just beyond the eyepiece.
Answer
Use the magnifier close to the eye; use short-focus lenses for higher magnification; place the eye at the eyepiece's exit pupil.
Question 26
A compound microscope must have angular magnification 30×. Its objective focal length is 1.25 cm and eyepiece focal length is 5 cm. Describe one suitable lens arrangement for normal adjustment, when the final image is at infinity.
Steps only
30=(L/1.25)(25/5), so L=1.5 cm between focal planes. Lens-centre separation=1.25+1.5+5=7.75 cm.
Answer
Separate lens centres by 7.75 cm; object about 1.56 cm before the objective.
Question 27
A small telescope has objective focal length 140 cm and eyepiece focal length 5 cm. Find the magnifying power when (a) the final image is at infinity and (b) the final image is at the near point 25 cm.
Steps only
At infinity M=140/5=28. At near point M=(140/5)(1+5/25)=33.6.
Answer
(a) 28×; (b) 33.6×, inverted.
Question 28
Use the telescope in Exercise 9.27(a) to observe a 100 m tower 3 km away. Find (a) the lens separation for a final image at infinity, (b) the image height formed by the objective, and (c) the final image height if adjusted for a final image at the 25 cm near point.
Steps only
Separation=145 cm; objective image=140×100/3000=4.67 cm. At the near point, eyepiece magnification=6, giving 28 cm.
Answer
145 cm; 4.67 cm at the objective; about 28 cm at the near point.
Question 29
A Cassegrain telescope has its primary concave mirror and secondary convex mirror 20 mm apart. The primary radius is 220 mm and the secondary radius is 140 mm. Where is the final image of an object at infinity?
Steps only
Primary focus is 110−20=90 mm beyond the secondary. For the convex secondary, f=70 mm and u=+90 mm: 1/v+1/90=1/70, so v=315 mm.
Answer
315 mm behind the secondary.
Question 30
A plane mirror in a current-sensitive galvanometer turns by 3.5° when current flows. A light spot is reflected onto a screen 1.5 m away. By how far does the spot move?
Steps only
Reflected-ray turn=2×3.5°=7°; y=1.5tan7°≈0.184 m.
Answer
About 18.4 cm.
Question 31
An equiconvex lens of refractive index 1.50 is placed on a liquid above a plane mirror. With the liquid present, a needle coincides with its inverted image at 45 cm from the lens. Without the liquid, the coincidence distance is 30 cm. Find the liquid’s refractive index.
Steps only
Without liquid, R=30 cm. With liquid, 1/45=(2−n_liquid)/30, so n_liquid=2−30/45=4/3.