Chapter 9: Ray Optics and Optical Instruments (Optics)

$f=-\frac{36}{2}=-18$ cm, $u=-27$ cm, $h=2.5$ cm.

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=-\frac{1}{54}\Rightarrow v=-54\text{cm}$$

The screen must be 54 cm in front of the mirror.

$m=-\frac{v}{u}=-\frac{-54}{-27}=-2$, so the image is $2\times 2.5=5$ cm tall, real, inverted and magnified.

As the candle moves closer (towards F), the image moves farther away, so the screen must be moved away from the mirror. Once the candle is inside the focus (closer than 18 cm), the image becomes virtual and cannot be caught on a screen.

Screen at 54 cm in front of the mirror; image real, inverted, 5 cm tall. Moving the candle closer needs the screen to move farther away (no screen image once inside F).

$f=+15$ cm, $u=-12$ cm.

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}\Rightarrow v=6.7\text{cm}$$

The image is 6.7 cm behind the mirror. $m=-\frac{v}{u}=-\frac{6.7}{-12}\approx 0.56$, so it is about $0.56\times 4.5=2.5$ cm tall, virtual, erect and diminished.

As the needle is moved farther away, the image moves towards the focus (but never beyond it) and keeps getting smaller.

6.7 cm behind the mirror, m ≈ 0.56 (2.5 cm, virtual, erect). Moving the needle away, the image shifts towards F and shrinks.

$$n=\frac{\text{real depth}}{\text{apparent depth}}=\frac{12.5}{9.4}\approx 1.33$$

With the new liquid, apparent depth $=\frac{12.5}{1.63}\approx 7.67$ cm. The needle now appears $9.4-7.67=1.73$ cm higher, so the microscope must be raised by about 1.7 cm.

n ≈ 1.33; the microscope is moved up by about 1.7 cm.

From the figures, the 60° ray refracts at 35° in glass and at 41° in water.

$$n_g=\frac{\sin 60^{\circ }}{\sin 35^{\circ }}=\frac{0.866}{0.574}\approx 1.51,n_w=\frac{\sin 60^{\circ }}{\sin 41^{\circ }}=\frac{0.866}{0.656}\approx 1.32$$

At the water-glass surface, $n_w\sin 45^{\circ }=n_g\sin r$:

$$\sin r=\frac{1.32\times 0.707}{1.51}\approx 0.618\Rightarrow r\approx 38^{\circ }$$

About 38°.

Light escapes only where it meets the surface at less than the critical angle. $\sin i_c=\frac{1}{1.33}=0.752$, so $i_c\approx 48.8^{\circ }$.

The light comes out through a circle of radius

$$r=h\tan i_c=0.80\times \tan 48.8^{\circ }=0.80\times 1.14\approx 0.91\text{m}$$

$$\text{Area}=\pi r^2=3.14\times (0.91{)}^2\approx 2.6{\text{m}}^2$$

About 2.6 m².

$$n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}=\frac{\sin 50^{\circ }}{\sin 30^{\circ }}=\frac{0.766}{0.5}\approx 1.53$$

In water, the refractive index of glass relative to water is $n^{\prime }=\frac{1.53}{1.33}\approx 1.15$:

$$\sin \frac{60^{\circ }+{\delta }_m^{\prime }}{2}=1.15\times \sin 30^{\circ }=0.576\Rightarrow \frac{60^{\circ }+{\delta }_m^{\prime }}{2}\approx 35.2^{\circ }\Rightarrow {\delta }_m^{\prime }\approx 10^{\circ }$$

n ≈ 1.53; in water the minimum deviation drops to about 10°.

For an equiconvex lens $R_1=R$, $R_2=-R$:

$$\frac{1}{f}=(n-1)\frac{2}{R}\Rightarrow R=2(n-1)f=2\times 0.55\times 20=22\text{cm}$$

R = 22 cm.

P acts as a virtual object on the far side of the lens: $u=+12$ cm.

(a) $f=+20$: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{12}=\frac{8}{60}$, so $v=7.5$ cm. The beam converges 7.5 cm from the lens.

(b) $f=-16$: $\frac{1}{v}=-\frac{1}{16}+\frac{1}{12}=\frac{1}{48}$, so $v=48$ cm. The beam converges 48 cm from the lens.

(a) 7.5 cm from the lens. (b) 48 cm from the lens (both on the side of P).

$f=-21$ cm, $u=-14$ cm:

$$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{21}-\frac{1}{14}=-\frac{5}{42}\Rightarrow v=-8.4\text{cm}$$

$m=\frac{v}{u}=\frac{-8.4}{-14}=0.6$, so the image is 1.8 cm tall: virtual, erect, diminished, 8.4 cm from the lens on the object's side.

As the object moves farther away, the image moves towards the focus (21 cm) and keeps getting smaller.

Virtual, erect image 1.8 cm tall, 8.4 cm from the lens on the object side; moving the object away, the image moves towards F and shrinks.

$$\frac{1}{F}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\Rightarrow F=-60\text{cm}$$

The negative focal length means the combination is a diverging lens.

F = −60 cm; diverging.

(a) Final image at 25 cm. Eyepiece: $v_e=-25$, $f_e=6.25$:

$\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25}=-0.2$, so $u_e=-5$ cm.

The objective's image is $15-5=10$ cm from it: $v_o=10$ cm. Then $\frac{1}{u_o}=\frac{1}{10}-\frac{1}{2}=-0.4$, so $u_o=-2.5$ cm.

$$m=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4\times 5=20$$

(b) Final image at infinity. The objective's image must be at the eyepiece's focus: $u_e=-6.25$ cm, so $v_o=15-6.25=8.75$ cm.

$\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2}=-0.386$, so $u_o\approx -2.59$ cm.

$$m=\frac{v_o}{|u_o|}\cdot \frac{D}{f_e}=\frac{8.75}{2.59}\times \frac{25}{6.25}\approx 13.5$$

(a) 2.5 cm from the objective, m = 20. (b) About 2.59 cm, m ≈ 13.5.

Objective: $f_o=0.8$ cm, $u_o=-0.9$ cm:

$\frac{1}{v_o}=\frac{1}{0.8}-\frac{1}{0.9}=0.139$, so $v_o=7.2$ cm.

Eyepiece (final image at the near point): $v_e=-25$, $f_e=2.5$:

$\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{2.5}=-0.44$, so $u_e\approx -2.27$ cm.

Separation $=7.2+2.27\approx 9.5$ cm.

$$m=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)=\frac{7.2}{0.9}\times \left(1+\frac{25}{2.5}\right)=8\times 11=88$$

Separation about 9.5 cm; magnifying power 88.

In normal adjustment: $m=\frac{f_o}{f_e}=\frac{144}{6.0}=24$, and the separation is $f_o+f_e=144+6=150$ cm.

m = 24; separation 150 cm.

(a) $m=\frac{f_o}{f_e}=\frac{1500\text{cm}}{1.0\text{cm}}=1500$.

(b) The moon subtends $\alpha =\frac{3.48\times 10^6}{3.8\times 10^8}\approx 9.16\times 10^{-3}$ rad. Its image at the focal plane of the objective has diameter

$$d=f_o\alpha =15\times 9.16\times 10^{-3}\approx 0.137\text{m}=13.7\text{cm}$$

(a) 1500. (b) About 13.7 cm.

The mirror equation is $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$. Real objects have $u<0$.

(a) Concave: $f=-F$ ($F>0$). Let $u=-x$ with $F<x<2F$. Then

$$\frac{1}{v}=-\frac{1}{F}+\frac{1}{x}=-\frac{x-F}{xF}\Rightarrow v=-\frac{xF}{x-F}$$

$v$ is negative, so the image is real (in front of the mirror). Its distance is $|v|=F\cdot \frac{x}{x-F}$. Because $x<2F$, we have $x>2(x-F)$, so $\frac{x}{x-F}>2$ and $|v|>2F$: the image is beyond 2f.

(b) Convex: $f=+F$. With $u=-x$: $\frac{1}{v}=\frac{1}{F}+\frac{1}{x}>0$, so $v>0$ for every object position: the image is always behind the mirror, virtual.

(c) From (b), $\frac{1}{v}>\frac{1}{F}$, so $v<F$: the image lies between the pole and the focus. Also $\frac{1}{v}>\frac{1}{x}$, so $v<x$ and $m=\frac{v}{x}<1$: diminished.

(d) Concave, $f=-F$, object inside the focus: $u=-x$ with $x<F$. Then $\frac{1}{v}=\frac{1}{x}-\frac{1}{F}>0$, so $v>0$: virtual. Also $v=\frac{xF}{F-x}$, so $m=\frac{v}{x}=\frac{F}{F-x}>1$: enlarged.

The signs of v and the size of m from the mirror equation give all four results.

A slab of thickness $t$ makes an object appear raised by

$$\Delta =t\left(1-\frac{1}{n}\right)=15\left(1-\frac{1}{1.5}\right)=5\text{cm}$$

This does not depend on where the slab is placed between the pin and the eye (for small angles of viewing).

Raised by 5 cm; independent of the slab's position.

(a) Critical angle at the core-cladding surface: $\sin i_c=\frac{1.44}{1.68}=0.857$, so $i_c\approx 59^{\circ }$.

If a ray inside makes angle $r$ with the axis, it meets the wall at $90^{\circ }-r$. Total internal reflection needs $90^{\circ }-r\ge 59^{\circ }$, so $r\le 31^{\circ }$.

At the entry face, $\sin i=1.68\sin r\le 1.68\sin 31^{\circ }\approx 0.865$, so $i\le 60^{\circ }$.

Rays entering at 0° to 60° with the axis are totally reflected.

(b) With air outside: $\sin i_c=\frac{1}{1.68}$, $i_c\approx 36.5^{\circ }$, so $r\le 53.5^{\circ }$. Then $1.68\sin 53.5^{\circ }\approx 1.35>1$, which means every ray that enters (any $i$ up to 90°) has $r$ below this limit. All incident rays suffer total internal reflection.

(a) 0° < i < 60° with the axis. (b) All angles of incidence (0° to 90°).

For a real image on a screen, the object-to-screen distance $D$ must be at least $4f$ (when $u=v=2f$). So $f\le \frac{D}{4}=\frac{3}{4}=0.75$ m.

0.75 m.

For the displacement method, with $D=90$ cm and $d=20$ cm, the two positions are $u=\frac{D\pm d}{2}$, giving $|u|=35$ cm, $v=55$ cm (and the reverse). So

$$f=\frac{D^2-d^2}{4D}=\frac{8100-400}{360}\approx 21.4\text{cm}$$

(Check: $\frac{1}{55}+\frac{1}{35}=\frac{90}{1925}$, $f=21.4$ cm.)

f ≈ 21.4 cm.

(a) Convex lens ($f=30$ cm) and concave lens ($f=-20$ cm), 8 cm apart.

Parallel light falling on the convex lens first: it would converge at 30 cm, which is 22 cm beyond the concave lens (virtual object, $u=+22$):

$\frac{1}{v}=-\frac{1}{20}+\frac{1}{22}=-\frac{1}{220}$, so $v=-220$ cm. The beam leaves diverging, as if from a point 220 cm before the concave lens, which is 216 cm from the centre of the system.

Parallel light falling on the concave lens first: it seems to come from its focus, 20 cm before it, i.e. 28 cm before the convex lens ($u=-28$):

$\frac{1}{v}=\frac{1}{30}-\frac{1}{28}=-\frac{1}{420}$, so $v=-420$ cm: diverging from a point 420 cm before the convex lens, which is 416 cm from the centre.

The answer depends on the side from which the light comes, so a single "effective focal length" is not a useful idea for this system.

(b) Convex lens: $u_1=-40$: $\frac{1}{v_1}=\frac{1}{30}-\frac{1}{40}=\frac{1}{120}$, $v_1=120$ cm, $m_1=\frac{120}{-40}=-3$.

This image is $120-8=112$ cm beyond the concave lens (virtual object, $u_2=+112$):

$\frac{1}{v_2}=-\frac{1}{20}+\frac{1}{112}=-\frac{23}{560}$, so $v_2\approx -24.3$ cm, $m_2=\frac{-24.3}{112}\approx -0.217$.

$m=m_1{m}_2=(-3)(-0.217)\approx 0.65$, so the image is $0.65\times 1.5\approx 0.98$ cm, virtual and erect, about 24.3 cm before the concave lens.

(a) It depends on the side (216 cm or 416 cm from the centre), so the idea is not useful. (b) m ≈ 0.65; image about 0.98 cm.

At the second face the angle must just equal the critical angle: $\sin r_2=\frac{1}{1.524}=0.656$, so $r_2=i_c\approx 41.0^{\circ }$.

Inside the prism $r_1+r_2=A$, so $r_1=60^{\circ }-41.0^{\circ }=19.0^{\circ }$.

At the first face: $\sin i=1.524\sin 19.0^{\circ }=1.524\times 0.326\approx 0.496$, so $i\approx 29.8^{\circ }$.

For any smaller angle of incidence, the ray is totally reflected at the second face.

About 29.8° (smaller angles give total internal reflection).

(a) The card is at the focus ($u=-9$ cm $=-f$), so $\frac{1}{v}=\frac{1}{9}-\frac{1}{9}=0$: the image is at infinity. The magnification $m=\frac{v}{u}$ is infinite, and so is the area of each square's image.

(b) Magnifying power $=\frac{D}{f}=\frac{25}{9}\approx 2.8$.

(c) No. Magnification compares the sizes of image and object; magnifying power compares the angles they subtend at the eye. With the image at infinity, the image is infinitely large but the angle it subtends is finite.

(a) Infinite (image at infinity), area infinite. (b) About 2.8. (c) No: one compares sizes, the other compares angles.

(a) Maximum magnifying power is obtained with the image at the near point, $v=-25$ cm:

$$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=-\frac{1}{25}-\frac{1}{9}=-\frac{34}{225}\Rightarrow u\approx -6.6\text{cm}$$

(b) $m=\frac{v}{u}=\frac{25}{6.6}\approx 3.8$.

(c) Yes. The magnifying power $1+\frac{D}{f}=1+\frac{25}{9}\approx 3.8$ equals the magnification here, because when the image is at the near point, the angle it subtends equals (image size)/D, and the object would have subtended (object size)/D.

(a) About 6.6 cm. (b) About 3.8. (c) Yes, when the image is at the near point (25 cm).

Area ratio 6.25 means linear magnification $m=\sqrt{6.25}=2.5$, so $v=2.5u$.

$$\frac{1}{2.5u}-\frac{1}{u}=\frac{1}{9}\Rightarrow \frac{-1.5}{2.5u}=\frac{1}{9}\Rightarrow u=-5.4\text{cm}$$

Then $v=-13.5$ cm. The image is only 13.5 cm from the eye, closer than the near point (25 cm), so you cannot see it distinctly with the eye close to the lens.

About 5.4 cm; no, the image (13.5 cm away) is nearer than 25 cm.

(a) Without the lens, we cannot see the object clearly closer than 25 cm, so the largest angle it can usefully subtend is (size)/25 cm. The magnifier lets us bring the object much closer than 25 cm and still see a clear image, so the angle at the eye is larger. That is the angular magnification.

(b) Yes, it decreases slightly. The angle subtended at the eye becomes a little smaller than the angle at the lens when the eye moves back. (The image also moves relative to the eye.)

(c) Lenses of very short focal length are hard to grind and suffer badly from spherical and chromatic aberration, so the image becomes poor. Also the object must be extremely close to the lens.

(d) $m=\frac{L}{f_o}\cdot \frac{D}{f_e}$: a large magnifying power needs both focal lengths small. A short $f_e$ also helps the eyepiece act as a strong magnifier of the objective's image.

(e) The eye should be at the eye-ring, the image of the objective formed by the eyepiece. All the light that passes through the objective passes through this small ring, so the eye collects all of it and sees the whole field of view. Its distance from the eyepiece depends on the separation of the lenses and on $f_e$; it is a short distance just beyond the eyepiece (roughly its focal length for a typical microscope).

(a) It lets the object be held closer than 25 cm, so the angle is larger. (b) It falls slightly. (c) Aberrations and practical limits. (d) Magnifying power is inversely proportional to both focal lengths. (e) At the eye-ring, just beyond the eyepiece, so all the light enters the eye.

Take the final image at the near point (25 cm). Eyepiece magnification: $m_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6$. So the objective must give $m_o=\frac{30}{6}=5$, i.e. $v_o=5|u_o|$.

Objective: $\frac{1}{v_o}+\frac{1}{|u_o|}=\frac{1}{f_o}$ gives $\frac{1}{5x}+\frac{1}{x}=\frac{1}{1.25}$, so $\frac{6}{5x}=0.8$ and $x=1.5$ cm. Then $u_o=-1.5$ cm and $v_o=7.5$ cm.

Eyepiece: $v_e=-25$, $\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}$, so $u_e\approx -4.17$ cm.

Separation $=7.5+4.17\approx 11.7$ cm.

Object 1.5 cm in front of the objective; lenses about 11.7 cm apart.

(a) $m=\frac{f_o}{f_e}=\frac{140}{5}=28$.

(b) $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=28\left(1+\frac{5}{25}\right)=33.6$.

(a) 28. (b) 33.6.

(a) Normal adjustment: $f_o+f_e=140+5=145$ cm.

(b) The tower subtends $\alpha =\frac{100}{3000}=\frac{1}{30}$ rad. The objective forms its image at its focal plane: height $=f_o\alpha =\frac{140}{30}\approx 4.7$ cm.

(c) The eyepiece magnifies this image by $1+\frac{D}{f_e}=1+\frac{25}{5}=6$, so the final image is $6\times 4.67=28$ cm tall.

(a) 145 cm. (b) About 4.7 cm. (c) 28 cm.

Large concave mirror: $f_1=\frac{220}{2}=110$ mm. Small convex mirror: $f_2=\frac{140}{2}=70$ mm.

Light from a distant object would form an image 110 mm from the large mirror. The small mirror is only 20 mm away, so this point lies $110-20=90$ mm behind the small mirror: a virtual object for it.

For the convex mirror (taking the direction of light reaching it as positive): $u=+90$ mm, $f=+70$ mm.

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{70}-\frac{1}{90}=\frac{2}{630}\Rightarrow v=315\text{mm}$$

The final image is real, 315 mm from the small mirror, on the side of the large mirror (it passes through the hole in the large mirror, about 295 mm behind it).

315 mm from the small mirror (about 295 mm behind the large mirror).

When a mirror turns by θ, the reflected ray turns by 2θ = 7°.

$$d=1.5\times \tan 7^{\circ }=1.5\times 0.123\approx 0.18\text{m}=18.4\text{cm}$$

About 18.4 cm.

The image coincides with the needle when the needle is at the focus of the lens system (the light returns along its own path from the mirror).

Without liquid: focal length of the glass lens $f_1=30$ cm. For an equiconvex lens, $\frac{1}{30}=0.5\times \frac{2}{R}$, so $R=30$ cm.

With liquid: the combination has $F=45$ cm. The liquid forms a plano-concave lens with one surface of radius 30 cm:

$$\frac{1}{f_2}=(\mu -1)\left(-\frac{1}{30}-0\right)=-\frac{\mu -1}{30}$$

$$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\Rightarrow \frac{1}{45}=\frac{1}{30}-\frac{\mu -1}{30}\Rightarrow \frac{\mu -1}{30}=\frac{1}{90}\Rightarrow \mu \approx 1.33$$

μ ≈ 1.33.