For a concave mirror, $f=-R/2=-18\,\text{cm}$ and $u=-27\,\text{cm}$. From $1/v+1/u=1/f$, $1/v=-1/18+1/27=-1/54$, so $v=-54\,\text{cm}$. The image is formed in front of the mirror and can be caught on a screen. Magnification $m=-v/u=-(-54)/(-27)=-2$, so image height $=-2(2.5)=-5.0\,\text{cm}$, i.e. inverted. As the object approaches the focus from beyond it, $|v|$ increases.
The screen should be placed $54\,\text{cm}$ in front of the mirror. The image is real, inverted and $5.0\,\text{cm}$ high. As the candle is moved closer to the mirror but remains beyond the focus, the screen must be moved farther away.
For a convex mirror, $f=+15\,\text{cm}$ and $u=-12\,\text{cm}$. From $1/v+1/u=1/f$, $1/v=1/15+1/12=9/60$, giving $v=6.67\,\text{cm}$. Magnification is $m=-v/u=-6.67/(-12)=0.56$. Positive $v$ means the image is behind the mirror; positive magnification means it is erect.
The image is $6.7\,\text{cm}$ behind the mirror, with magnification $0.56$. It is virtual, erect and diminished. As the needle is moved farther away, the image moves towards the focus and becomes smaller.
For near-normal viewing, $\mu=\text{real depth}/\text{apparent depth}=12.5/9.4=1.33$. In the new liquid, apparent depth $=12.5/1.63=7.67\,\text{cm}$. The image appears closer to the surface than before by $9.4-7.67=1.73\,\text{cm}$, so the microscope must be moved upward by this amount.
The refractive index of water is $1.33$. With the liquid of refractive index $1.63$, the microscope must be raised by about $1.7\,\text{cm}$.
Light emerges only within the cone whose semi-angle is the critical angle. For water-air, $\sin i_c=1/1.33$, so $\tan i_c=1.14$. The radius of the emergent circular patch is $r=h\tan i_c=0.80(1.14)=0.91\,\text{m}$. Area $=\pi r^2=2.6\,\text{m}^2$.
$2.6\,\text{m}^2$ approximately.
For minimum deviation, $n=\sin[(A+D_m)/2]/\sin(A/2)=\sin50^\circ/\sin30^\circ=1.53$. In water, the relative refractive index is $1.53/1.33=1.15$. Thus $\sin[(60^\circ+D'_m)/2]=1.15\sin30^\circ=0.576$. Hence $(60^\circ+D'_m)/2=35.2^\circ$ and $D'_m=10.4^\circ$.
The refractive index of the prism is $1.53$. In water, the new angle of minimum deviation is about $10.4^\circ$.
For an equiconvex lens in air, $R_1=R$, $R_2=-R$, and $1/f=(n-1)(1/R_1-1/R_2)=2(n-1)/R$. Therefore $R=2(n-1)f=2(0.55)(20)=22\,\text{cm}$.
$22\,\text{cm}$.
The point P acts as a virtual object for the lens, so $u=+12\,\text{cm}$. Using $1/v-1/u=1/f$: for the convex lens, $1/v=1/20+1/12$, so $v=7.5\,\text{cm}$. For the concave lens, $f=-16\,\text{cm}$, so $1/v=-1/16+1/12=1/48$, giving $v=48\,\text{cm}$.
(a) $7.5\,\text{cm}$ beyond the lens. (b) $48\,\text{cm}$ beyond the lens.
For a concave lens, $f=-21\,\text{cm}$ and $u=-14\,\text{cm}$. From $1/v-1/u=1/f$, $1/v=-1/21-1/14=-5/42$, so $v=-8.4\,\text{cm}$. Magnification $m=v/u=(-8.4)/(-14)=0.60$, so image height $=0.60(3.0)=1.8\,\text{cm}$.
The image is virtual, erect and diminished, formed $8.4\,\text{cm}$ from the lens on the object side, with height $1.8\,\text{cm}$. If the object is moved farther away, the image moves towards the focus and becomes still smaller.
For lenses in contact, $1/F=1/f_1+1/f_2=1/30+1/(-20)=-1/60$. Thus $F=-60\,\text{cm}$, so the combination behaves as a diverging lens.
$-60\,\text{cm}$; the system is diverging.
For final image at $25\,\text{cm}$, the eyepiece has $v_e=-25\,\text{cm}$ and $f_e=6.25\,\text{cm}$, giving $u_e=-5\,\text{cm}$. Thus the objective image is $10\,\text{cm}$ from the objective. With $f_o=2\,\text{cm}$, $1/10-1/u_o=1/2$, so $u_o=-2.5\,\text{cm}$. Objective magnification is $4$ in magnitude and eyepiece magnification is $1+D/f_e=5$, so total is $20$. For final image at infinity, $u_e=-6.25\,\text{cm}$, so objective image distance is $8.75\,\text{cm}$. Then $u_o=-2.59\,\text{cm}$. The magnifying power is $(8.75/2.59)(25/6.25)=13.5$ approximately.
(a) Object distance $2.5\,\text{cm}$; magnifying power $20$. (b) Object distance $2.59\,\text{cm}$; magnifying power about $13.5$.
For the objective, $f_o=0.80\,\text{cm}$ and $u_o=-0.90\,\text{cm}$. From $1/v_o-1/u_o=1/f_o$, $v_o=7.2\,\text{cm}$. For final image at the near point, the eyepiece has $f_e=2.5\,\text{cm}$ and $v_e=-25\,\text{cm}$, giving $u_e=-2.27\,\text{cm}$. Hence lens separation $=7.2+2.27=9.47\,\text{cm}$. Objective magnification is $|v_o/u_o|=8$ and eyepiece magnification is $1+25/2.5=11$, so total magnifying power $=88$.
The lens separation is about $9.5\,\text{cm}$ and the magnifying power is about $88$.
For normal adjustment, magnifying power of a telescope is $m=f_o/f_e=144/6.0=24$. The lens separation is $f_o+f_e=144+6.0=150\,\text{cm}$.
Magnifying power $=24$; separation $=150\,\text{cm}$.
Angular magnification in normal adjustment is $f_o/f_e=15/0.010=1500$. The angular diameter of the moon is approximately $3.48\times10^6/(3.8\times10^8)=9.16\times10^{-3}\,\text{rad}$. The objective image diameter is $f_o\theta=15(9.16\times10^{-3})=0.137\,\text{m}$.
(a) $1500$. (b) $0.137\,\text{m}$, or about $13.7\,\text{cm}$.
For a concave mirror, $f<0$ and a real object has $u<0$. If $f<u<2f$ in magnitude, solving the mirror equation gives $v<2f$ in sign convention, meaning a real image beyond $2f$. For a convex mirror, $f>0$ and $u<0$, so $1/v=1/f-1/u$ is positive for all object positions; hence $v>0$, a virtual image behind the mirror. Also $v<f$, so it lies between pole and focus, and $m=-v/u<1$, so it is diminished. For a concave mirror with the object between pole and focus, $|u|<|f|$, which gives $v>0$, a virtual image; $m=-v/u>1$, so it is enlarged.
All four results follow directly from $1/v+1/u=1/f$ with the Cartesian sign convention.
Apparent upward shift through a parallel slab is $t(1-1/\mu)$. Thus shift $=15[1-1/1.5]=15(1/3)=5\,\text{cm}$. The result depends on thickness and refractive index, not on where the slab is placed between eye and object.
The pin appears raised by $5\,\text{cm}$. The shift does not depend on the slab's location, as long as it is parallel to the table.
For a real image on a screen separated from the object by fixed distance $D=3\,\text{m}$, the maximum focal length occurs when object and image distances are equal: $u=v=D/2$. Then $f=uv/(u+v)=D/4=3/4=0.75\,\text{m}$.
$0.75\,\text{m}$.
Using the displacement method, $f=(D^2-d^2)/(4D)$, where $D=90\,\text{cm}$ is the object-screen distance and $d=20\,\text{cm}$ is the separation between lens positions. Thus $f=(90^2-20^2)/(4\times90)=7700/360=21.4\,\text{cm}$.
$21.4\,\text{cm}$.
For just total internal reflection at the second face, the internal incidence angle there equals the critical angle, $\sin c=1/1.524$, so $c=41.0^\circ$. Inside the prism, $r_1+r_2=A=60^\circ$, so $r_1=60^\circ-41.0^\circ=19.0^\circ$. At the first face, $\sin i=1.524\sin19.0^\circ=0.496$, giving $i=29.8^\circ$.
$29.8^\circ$ approximately.
The card is at the focal plane of the magnifying glass, so the virtual image is formed at infinity. Therefore its linear size cannot be assigned a finite value. The angular magnification for relaxed viewing is $D/f=25/9=2.8$. This is not the same as linear magnification; angular magnification compares the viewing angle with and without the magnifier.
(a) The image is at infinity, so linear magnification and image area are not finite useful quantities. (b) Angular magnification is $25/9=2.8$. (c) No; linear magnification and angular magnification are different concepts.
For maximum magnifying power, the final image is at the near point, $v=-25\,\text{cm}$, with $f=9\,\text{cm}$. From $1/v-1/u=1/f$, $-1/25-1/u=1/9$, giving $u=-6.62\,\text{cm}$. The linear magnification is $|v/u|=25/6.62=3.8$, equal to $1+D/f=1+25/9=3.8$ when the eye is close to the lens.
(a) $6.6\,\text{cm}$. (b) $3.8$. (c) With the eye close to the lens and final image at the near point, the linear magnification equals the angular magnifying power numerically.
Area magnification is $6.25/1=6.25$, so linear magnification is $m=2.5$. With $m=v/u=2.5$ and $1/v-1/u=1/f$ for $f=9\,\text{cm}$, putting $v=2.5u$ gives $u=-5.4\,\text{cm}$ and $v=-13.5\,\text{cm}$. Since the final image is at $13.5\,\text{cm}$, it is closer than the least distance of distinct vision, so it cannot be seen distinctly by a normal eye close to the lens.
The object should be $5.4\,\text{cm}$ from the lens. No; the virtual image would be only $13.5\,\text{cm}$ from the lens, closer than the normal near point.
Using $m=(L/f_o)(1+D/f_e)$ with $D=25\,\text{cm}$, $f_o=1.25\,\text{cm}$ and $f_e=5\,\text{cm}$, $30=(L/1.25)(1+25/5)=6L/1.25$, so $L=6.25\,\text{cm}$, where $L$ is the distance between the second focal point of the objective and the first focal point of the eyepiece. Thus lens separation $=L+f_o+f_e=12.5\,\text{cm}$. For final image at $25\,\text{cm}$, the eyepiece object distance is $4.17\,\text{cm}$, so the objective image distance is $12.5-4.17=8.33\,\text{cm}$. The objective formula gives $u_o=-1.47\,\text{cm}$.
For final image at the near point, set the lens separation to about $12.5\,\text{cm}$ and place the object about $1.47\,\text{cm}$ in front of the objective.
For normal adjustment, $m=f_o/f_e=140/5.0=28$. When the final image is at the near point, $m=(f_o/f_e)(1+f_e/D)=28(1+5/25)=33.6$.
(a) $28$. (b) $33.6$.
In normal adjustment, separation is $f_o+f_e=140+5=145\,\text{cm}$. The tower subtends angle $\theta=h/d=100/3000=1/30\,\text{rad}$. The objective image height is $f_o\theta=140\,\text{cm}/30=4.67\,\text{cm}$. If the final image is at $25\,\text{cm}$, eyepiece magnification is $1+D/f_e=1+25/5=6$, so final image height is $6(4.67)=28\,\text{cm}$.
(a) $145\,\text{cm}$. (b) $4.7\,\text{cm}$. (c) About $28\,\text{cm}$.
When the mirror turns by $3.5^\circ$, the reflected ray turns by twice this angle, $7.0^\circ$. The spot displacement on a screen at distance $L=1.5\,\text{m}$ is $d=L\tan7.0^\circ=1.5(0.123)=0.184\,\text{m}$.
$0.18\,\text{m}$ approximately.