Solution: Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. Differentiate w.r.t time $t$: $-\frac{1}{v^2}\frac{dv}{dt} - \frac{1}{u^2}\frac{du}{dt} = 0$.
Velocity of image $v_i = \frac{dv}{dt}$. Velocity of object $v_o = \frac{du}{dt}$.
$-\frac{v_i}{v^2} = \frac{v_o}{u^2} \Rightarrow v_i = -\left(\frac{v}{u}\right)^2 v_o$. Since $m = -\frac{v}{u}$, we have $v_i = -m^2 v_o$. The negative sign indicates the image always moves in the opposite direction to the object relative to the mirror.
Solution: Let object ends be at $u_1, u_2$. Length $L = du = u_2 - u_1$. Image ends are at $v_1, v_2$. Length $dL = dv = v_2 - v_1$. From mirror formula differentiation: $dv = -\left(\frac{v}{u}\right)^2 du$. Longitudinal magnification $m_L = \left|\frac{dv}{du}\right| = \left(\frac{v}{u}\right)^2$. Since transverse magnification $m = -v/u$, we get $m_L = m^2$.
Solution: Object is at $u_1 = -1.5f$ from $M_1$. Reflection 1 at $M_1$: $\frac{1}{v_1} = \frac{1}{-f} - \frac{1}{-1.5f} = -\frac{1}{f} + \frac{2}{3f} = \frac{-3+2}{3f} = -\frac{1}{3f}$. So $v_1 = -3f$. The first image $I_1$ forms exactly at the pole of $M_2$ (since $d=3f$).
Reflection 2 at $M_2$: Object $I_1$ is at $u_2 = 0$ for $M_2$. Image forms at $v_2 = 0$. So the second image $I_2$ forms exactly at the pole of $M_2$, which is at a distance of $3f$ from $M_1$.
Solution: The two halves act as independent mirrors. Object distance $u = -30\text{cm}$, $f = -20\text{cm}$. $\frac{1}{v} = \frac{1}{-20} - \frac{1}{-30} = \frac{-3+2}{60} = -\frac{1}{60} \Rightarrow v = -60\text{cm}$. Magnification $m = -(-60)/(-30) = -2$.
When pulled apart by $1\text{cm}$, the principal axes of the two halves are separated by $1\text{cm}$. The object is now $0.5\text{cm}$ below the axis of the upper half and $0.5\text{cm}$ above the axis of the lower half.
For upper half: $h = -0.5\text{cm} \Rightarrow h' = m \times h = (-2)(-0.5) = +1\text{cm}$ (above upper axis).
For lower half: $h = +0.5\text{cm} \Rightarrow h' = m \times h = (-2)(0.5) = -1\text{cm}$ (below lower axis).
Total distance = Distance between axes + $h'_{upper} + |h'_{lower}|$? No, use absolute coordinates from original axis. Upper axis is at $y = +0.5$. Image is at $y = 0.5 + 1 = 1.5\text{cm}$. Lower axis is at $y = -0.5$. Image is at $y = -0.5 - 1 = -1.5\text{cm}$. Distance between images $= 1.5 - (-1.5) = 3.0\text{ cm}$.
Solution:
AI Prompt: Create a mathematically correct physics ray diagram showing spherical aberration in a concave mirror of large aperture. Show parallel incident rays. The paraxial rays (close to axis) converge at the true focus F. The marginal rays (far from axis) reflect and converge at a point closer to the mirror pole than F. Fully white background, landscape mode, precise clean lines.
File Name: Level3_Q5_SphericalAberration.png
Solution: The system acts as a combination of a plano-convex liquid lens and a concave mirror. Focal length of water lens: $\frac{1}{f_l} = (\mu-1)(\frac{1}{\infty} - \frac{1}{-R}) = \frac{\mu-1}{R}$. Focal length of mirror $f_m = -R/2$. Equivalent focal length $F$: $P = P_l + P_m + P_l$ (ray passes lens, hits mirror, passes lens again).
$P_{eq} = -\frac{1}{F} = 2(\frac{1}{f_l}) - \frac{1}{f_m} = 2\left(\frac{\mu-1}{R}\right) - \left(-\frac{2}{R}\right) = \frac{2\mu - 2 + 2}{R} = \frac{2\mu}{R}$. Thus, $F = -\frac{R}{2\mu}$.
Solution: Let $u = -(f + x_1)$ and $v = -(f + x_2)$. Substitute into mirror formula: $\frac{1}{-f} = \frac{1}{-(f+x_2)} + \frac{1}{-(f+x_1)}$. $\Rightarrow \frac{1}{f} = \frac{1}{f+x_2} + \frac{1}{f+x_1} = \frac{f+x_1 + f+x_2}{(f+x_1)(f+x_2)}$.
$(f+x_1)(f+x_2) = f(2f + x_1 + x_2) \Rightarrow f^2 + f x_2 + x_1 f + x_1 x_2 = 2f^2 + f x_1 + f x_2$.
Canceling terms gives $x_1 x_2 = f^2$.
Solution: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$. Let $y = 1/v$, $x = 1/u$. Equation is $y = -x + c$, where $c = 1/f$. Slope $m = -1$. Intercept on y-axis (when $1/u=0$) is $1/f$. Intercept on x-axis (when $1/v=0$) is $1/f$.
Solution: $f = \frac{uv}{u+v} = \frac{30 \times 60}{30+60} = \frac{1800}{90} = 20.0\text{ cm}$.
Error formula from $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$.
$\Delta f = f^2 \left( \frac{0.2}{3600} + \frac{0.1}{900} \right) = 400 \left( \frac{0.2}{3600} + \frac{0.4}{3600} \right) = 400 \left( \frac{0.6}{3600} \right) = \frac{0.6}{9} \approx 0.067\text{ cm}$.
Percentage error $= (\Delta f / f) \times 100 = (0.067 / 20) \times 100 \approx 0.33\%$.
Solution: The mirror reflects a converging cone of light that meets at distance $v$. After passing $v$, the cone diverges. Let the image be at $v$. The distance from the image to the plane at $x$ is $(x-v)$. By similar triangles, the ratio of the diameter $d$ of the beam at $x$ to the mirror diameter $D$ is equal to the ratio of their distances from the image: $\frac{d}{x-v} = \frac{D}{v} \Rightarrow d = D \frac{x-v}{v}$. Where $v = \frac{uf}{u-f}$.
Solution: Snell's law in variable medium: $n(y) \sin i = \text{constant}$. At origin, $y=0$, $n=n_0$, $i=\theta$. Constant $= n_0 \sin\theta$. At any point $(x,y)$, angle with normal (y-axis) is $\alpha$. $n(y) \sin\alpha = n_0 \sin\theta$. From geometry of curve, $\sin\alpha = \frac{dx}{\sqrt{dx^2+dy^2}} = \frac{1}{\sqrt{1+(dy/dx)^2}}$.
$n_0 \sqrt{1+ky} \frac{1}{\sqrt{1+(dy/dx)^2}} = n_0 \sin\theta \Rightarrow \frac{1+ky}{1+(dy/dx)^2} = \sin^2\theta \Rightarrow 1+(dy/dx)^2 = \frac{1+ky}{\sin^2\theta}$.
$(dx/dy)^2 = \frac{\sin^2\theta}{1+ky - \sin^2\theta} = \frac{\sin^2\theta}{\cos^2\theta + ky}$. Integrating this yields the parabolic trajectory equation.
Solution: Apparent depth is additive for layers viewed normally. For layer 1, apparent depth shift is $d_1/\mu_1$. Total apparent depth $D_{app} = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} + \frac{d_3}{\mu_3} = \sum \frac{d_i}{\mu_i}$.
Solution: Let bird's real height be $y$, fish's depth be $x$. Apparent height of bird to fish is $y' = n_{water} \times y = \frac{4}{3}y$. The apparent velocity of bird is $\frac{dy'}{dt} = \frac{4}{3}\frac{dy}{dt} = \frac{4}{3}(2) = \frac{8}{3}\text{ m/s}$ (downwards). The fish is moving upwards at $1\text{ m/s}$. Relative velocity $= V_{bird,apparent} + V_{fish} = \frac{8}{3} + 1 = \frac{11}{3}\text{ m/s}$.
Solution: Ray enters normally through short face, hits hypotenuse at $45^\circ$. For TIR, $i \geq C$. Here $i = 45^\circ$. So $45^\circ \geq C \Rightarrow \sin 45^\circ \geq \sin C$. $\sin C = n_{liquid} / n_{glass}$. $1/\sqrt{2} \geq n_L / 1.5 \Rightarrow n_L \leq 1.5 / \sqrt{2} = 1.5 / 1.414 \approx 1.06$. Max refractive index of liquid is $1.06$.
Solution: Let incident angle be $\theta_i$. Refracted angle in core $r$. $\sin\theta_i = n_1 \sin r$. At core-cladding, $i' = 90^\circ - r$. For TIR, $\sin i' \geq n_2/n_1 \Rightarrow \cos r \geq n_2/n_1$. Using $\sin^2 r = 1 - \cos^2 r \leq 1 - (n_2/n_1)^2 = \frac{n_1^2 - n_2^2}{n_1^2}$. Thus, $\sin\theta_i = n_1 \sin r \leq n_1 \sqrt{\frac{n_1^2 - n_2^2}{n_1^2}} = \sqrt{n_1^2 - n_2^2}$. The maximum value of $\sin\theta_i$ is the Numerical Aperture: $NA = \sqrt{n_1^2 - n_2^2}$. Max acceptance angle $\theta_{max} = \sin^{-1}(\sqrt{n_1^2 - n_2^2})$.
Solution:
AI Prompt: Create a mathematically correct physics ray diagram showing the acceptance cone of an optical fiber. Show the core and cladding. Draw a 3D-like cone at the entrance face representing the maximum angle of incidence (acceptance angle) for which rays will undergo Total Internal Reflection inside the core. Label n_1, n_2, and the acceptance angle theta_a. Fully white background, landscape mode.
File Name: Level3_Q16_AcceptanceCone.png
Solution: Lateral shift $x = \frac{t \sin(i-r)}{\cos r}$. This is maximum when incidence is grazing, $i \to 90^\circ$. Then $\sin(90^\circ-r) / \cos r = \cos r / \cos r = 1$. So $x_{max} = t$. The ray travels along the surface.
Solution: From opposite side, object distance $u = -(R + R/2) = -1.5R$. Light travels from $n_1=1.5$ to $n_2=1$. Surface radius is $-R$. $\frac{1}{v} - \frac{1.5}{-1.5R} = \frac{1 - 1.5}{-R} \Rightarrow \frac{1}{v} + \frac{1}{R} = \frac{0.5}{R} \Rightarrow \frac{1}{v} = \frac{0.5}{R} - \frac{1}{R} = -\frac{0.5}{R} = -\frac{1}{2R}$. Thus $v = -2R$. It appears at a distance of $2R$ from the surface.
Solution: The system acts as a mirror of equivalent focal length. Power of system $P = P_L + P_M + P_L$. Alternatively, track the ray: (1) Refraction at first surface: object appears shifted to $u' = 15 \times 1.5 = 22.5\text{cm}$ from surface, so $22.5+10=32.5\text{cm}$ from silvered mirror. (2) Mirror reflects it to $32.5\text{cm}$ in front of mirror. (3) Ray travels $10\text{cm}$ through glass to front surface, arriving at $32.5 - 10 = 22.5\text{cm}$ apparent depth from inside. (4) Final refraction into air: apparent depth $= 22.5 / 1.5 = 15\text{cm}$. Image forms exactly at the object position ($15\text{cm}$ in front of the slab).
Solution: Cauchy's formula: $n = A + B/\lambda^2$. As wavelength $\lambda$ increases (e.g., from violet to red), refractive index $n$ decreases. Since $\sin C = 1/n$, a decrease in $n$ leads to an increase in $\sin C$, and thus an increase in the critical angle $C$. Red light has a larger critical angle than violet light.
Solution: System is Lens + Mirror + Lens. Power $P = P_L + P_M + P_L = 2P_L + P_M$. $P_L = \frac{1}{f_l} = (\mu-1)(\frac{2}{R})$ (assuming equiconvex). $P_M = -\frac{1}{f_m} = -\frac{1}{-R/2} = \frac{2}{R}$. $P = 2(\mu-1)\frac{2}{R} + \frac{2}{R} = \frac{4\mu - 4 + 2}{R} = \frac{4\mu-2}{R}$. Equivalent focal length $F = -1/P = -\frac{R}{4\mu-2}$. It acts as a concave mirror.
Solution: Ray parallel to axis hits $L_1$ at height $h_1$. Deviation $\delta_1 = h_1/f_1$. It hits $L_2$ at height $h_2 = h_1 - \delta_1 d = h_1 - h_1 d / f_1$. Deviation at $L_2$ is $\delta_2 = h_2/f_2$. Total deviation $\delta = \delta_1 + \delta_2 = \frac{h_1}{f_1} + \frac{h_1(1-d/f_1)}{f_2}$. Equivalent lens at height $h_1$ causes same deviation: $\delta = h_1/F$. $\frac{h_1}{F} = h_1(\frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2})$. Thus $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$.
Solution: $\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0$. Since dispersive powers $\omega_1, \omega_2$ are always positive, the only way their sum can be zero is if $f_1$ and $f_2$ have opposite signs. Thus, one lens must be convex (positive $f$) and the other must be concave (negative $f$).
Solution: In the two conjugate positions, $u_1 = v_2$ and $v_1 = u_2$. Magnifications are $m_1 = v_1/u_1$ and $m_2 = v_2/u_2 = u_1/v_1 = 1/m_1$. So $m_1 m_2 = 1$. Also $m_1 = I_1/O$ and $m_2 = I_2/O$. Therefore $(I_1/O)(I_2/O) = 1 \Rightarrow O^2 = I_1 I_2 \Rightarrow O = \sqrt{I_1 I_2}$.
Solution: The two halves have the same focal length $f$. They form images at the same horizontal distance $v$ given by $1/v - 1/u = 1/f$. However, their principal axes are shifted by $a$. Magnification $m = v/u$. For upper half, object is at $y = -a/2$ relative to its axis. Image $y'_1 = m(-a/2)$. For lower half, object is at $y = +a/2$. Image $y'_2 = m(+a/2)$. Distance between images relative to their respective axes is $ma$. Since axes are $a$ apart, total distance $= a + |ma|$? Actually, using coordinate geometry, distance is $a(1 + v/|u|)$ or $a(1-m)$. This setup creates two coherent sources (Billet's split lens).
Solution:
AI Prompt: Create a mathematically correct physics ray diagram showing an achromatic doublet. Show a biconvex lens (Crown glass) cemented to a plano-concave lens (Flint glass). Draw parallel rays of white light incident on the doublet, passing through, and converging precisely at a single focal point, demonstrating the elimination of longitudinal chromatic aberration. Fully white background, landscape mode.
File Name: Level3_Q26_AchromaticDoublet.png
Solution: $1/f \propto (\frac{n_g}{n_l} - 1)$. As $T$ increases, $n_l$ decreases much faster than $n_g$, so the ratio $n_g/n_l$ increases. Therefore, the term $(n_g/n_l - 1)$ increases, which means $1/f$ increases. Consequently, the focal length $f$ decreases over time.
Solution: System is 3 lenses: $L_1$ (concavo-convex glass), $L_2$ (biconvex water), $L_3$ (convexo-concave glass).
$1/f_1 = (1.5-1)(-1/20 - (-1/10)) = 0.5(1/20) = +1/40$.
$1/f_2 = (1.33-1)(1/10 - (-1/10)) = (1/3)(2/10) = +1/15$.
$1/f_3 = (1.5-1)(1/10 - 1/20) = 0.5(1/20) = +1/40$.
$1/F = 1/40 + 1/15 + 1/40 = 2/40 + 1/15 = 1/20 + 1/15 = (3+4)/60 = 7/60 \Rightarrow F = 60/7 \approx 8.57\text{ cm}$.
Solution: Lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Differentiate w.r.t $t$: $-\frac{1}{v^2} v_i + \frac{1}{u^2} v_o = 0 \Rightarrow v_i = (\frac{v}{u})^2 v_o = m^2 v_o$.
Substitute $v = \frac{uf}{u+f}$, then $v/u = \frac{f}{u+f}$. Therefore, $v_i = \left(\frac{f}{u+f}\right)^2 v_o$. Both move in same direction relative to space.
Solution: $\frac{1}{f} = (n-1) \left[ \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)t}{n R_1 R_2} \right]$.
Solution: For no emergence, $r_2 \geq C$ always. We know $r_1 + r_2 = A \Rightarrow r_2 = A - r_1$. So $A - r_1 \geq C$. To ensure this is true for *all* $i$, it must be true for the maximum possible $r_1$. Max $i = 90^\circ \Rightarrow \max r_1 = C$. So $A - C \geq C \Rightarrow A \geq 2C$. If the prism angle is greater than twice the critical angle, no ray can emerge from the second face.
Solution: Condition for zero mean deviation: $\delta_1 + \delta_2 = 0 \Rightarrow (\mu_{y1}-1)A_1 + (\mu_{y2}-1)A_2 = 0 \Rightarrow A_1/A_2 = -(\mu_{y2}-1)/(\mu_{y1}-1)$.
Net dispersion $\theta = \theta_1 + \theta_2 = (\mu_{v1}-\mu_{r1})A_1 + (\mu_{v2}-\mu_{r2})A_2 = \omega_1 \delta_1 + \omega_2 \delta_2$. Since $\delta_2 = -\delta_1$, net dispersion $\theta = \delta_1(\omega_1 - \omega_2)$.
Solution: To retrace its path, the ray must strike the silvered back face normally. So $r_2 = 0$. Since $r_1 + r_2 = A$, $r_1 = A = 6^\circ$. By Snell's Law at first face: $\sin i = \mu \sin r_1 \approx \mu r_1$ (for small angles). $i = 1.5 \times 6^\circ = 9^\circ$.
Solution:
AI Prompt: Create a mathematically correct physics ray diagram showing two thin prisms inverted relative to each other (e.g., Crown glass followed by Flint glass). A white light beam enters the first prism, disperses, then enters the second prism. The second prism cancels the mean deviation so the emergent yellow ray is parallel to the incident ray, but the red and violet rays remain angularly separated (dispersion without deviation). Fully white background, landscape mode.
File Name: Level3_Q34_DispersionNoDeviation.png
Solution: Deviation $\delta = i + e - A$. $\delta$ is maximum when $i = 90^\circ$ (grazing incidence). In this case, $r_1 = C$. Then $r_2 = A - C$. $e = \sin^{-1}(\mu \sin(A-C))$. Max deviation $\delta_{max} = 90^\circ + \sin^{-1}(\mu \sin(A-C)) - A$.
Solution: By Cauchy's formula, $\mu$ decreases as $\lambda$ increases. Since $\delta_m \approx (\mu-1)A$ (or the exact formula is monotonically dependent on $\mu$), $\delta_m$ decreases as $\lambda$ increases. The graph is a decreasing curve (higher deviation for violet, lower for red).
Solution: Symmetrical means minimum deviation, so $r = A/2$. $n = \frac{\sin i}{\sin r} = \frac{\sin(3A/4)}{\sin(A/2)}$.
Solution: Equilateral means $A = 60^\circ$. $\delta_m = 30^\circ$. $n = \frac{\sin((60+30)/2)}{\sin(60/2)} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414$.
Solution: Dispersive power $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$. From $n = a + b/\lambda^2$, $dn/d\lambda = -2b/\lambda^3$. This shows the change in refractive index is highly non-linear with respect to wavelength. The dispersion spread between violet and blue is greater than between red and orange. Therefore, $\omega$ depends precisely on which specific wavelengths are chosen to define the "extremes" and the "mean".
Solution: $(A+\delta_m)/2 = (60^\circ + 39^\circ 30')/2 = 99^\circ 30' / 2 = 49^\circ 45'$.
$n = \frac{\sin 49^\circ 45'}{\sin 30^\circ} = \frac{0.763}{0.5} = 1.526$.
Solution: Resolving power is the reciprocal of the limit of resolution ($d$), which is the minimum distance between two points to be seen distinctly. $RP = 1/d = \frac{2 n \sin\theta}{1.22 \lambda} = \frac{2 NA}{1.22 \lambda}$. Higher Numerical Aperture (NA) increases resolving power.
Solution: Rayleigh criterion: Two point sources are just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other. The angular radius of the first minimum for a circular aperture $D$ is $\theta = 1.22 \lambda / D$. Resolving power is $1/\Delta\theta = D / 1.22 \lambda$.
Solution:
AI Prompt: Create a mathematically correct physics ray diagram of a Galilean telescope in normal adjustment. Show a convex objective lens and a concave eyepiece lens. Parallel rays from infinity enter the objective, converge towards its focus, but are intercepted by the concave eyepiece placed before the focal point. The rays exit the concave eyepiece as parallel rays. The background of the whole image should be fully white. It should be in landscape mode. High quality, precise clean lines.
File Name: Level3_Q43_GalileanTelescope.png
Solution: $M_1 = \frac{v_o}{u_o} (\frac{D}{f_e})$. $M_2 = \frac{v_o}{u_o} (1 + \frac{D}{f_e})$. Ratio $\frac{M_2}{M_1} = \frac{1 + D/f_e}{D/f_e} = \frac{f_e + D}{D} = 1 + \frac{f_e}{D}$.
Solution: The erecting lens $f$ does not change magnification, it only inverts the image. $M = f_o / f_e$. Tube length $L = f_o + 4f + f_e$ (since object is at $2f$ and image at $2f$ for the erecting lens).
Solution: $M = f_o / f_e = 10 \Rightarrow f_o = 10 f_e$. $L = f_o + f_e = 44$. $10 f_e + f_e = 44 \Rightarrow 11 f_e = 44 \Rightarrow f_e = 4\text{ cm}$. $f_o = 10(4) = 40\text{ cm}$.
Solution: Distant vision (upper part): Needs to bring $\infty$ to $300\text{cm}$. $u = -\infty, v = -300\text{cm}$. $1/f = 1/-3 - 1/-\infty = -1/3\text{m} \Rightarrow P_{upper} = -0.33\text{ D}$.
Near vision (lower part): Needs to bring $25\text{cm}$ to $50\text{cm}$. $u = -0.25\text{m}, v = -0.50\text{m}$. $1/f = 1/-0.50 - 1/-0.25 = -2 + 4 = +2\text{m}^{-1} \Rightarrow P_{lower} = +2.0\text{ D}$.
Solution: Angle by object $\alpha = h / u_o$. Objective forms image $h'$ at $v_o$, so $h'/h = v_o/u_o \Rightarrow \alpha = h'/v_o$. Angle by final image $\beta = h'/u_e$. $M = \beta/\alpha = (h'/u_e) / (h'/v_o) = v_o / u_e$. For eyepiece, $1/v_e - 1/u_e = 1/f_e \Rightarrow 1/-D - 1/-u_e = 1/f_e \Rightarrow 1/u_e = 1/f_e + 1/D = \frac{1}{f_e}(1 + f_e/D)$. Thus $M = \frac{v_o}{f_e} (1 + \frac{f_e}{D})$.
Solution: The exit pupil is the image of the objective lens formed by the eyepiece. Let objective diameter be $D_{obj}$. Treating objective as object for eyepiece, $u = -(f_o + f_e)$. $1/v - 1/-(f_o+f_e) = 1/f_e \Rightarrow 1/v = 1/f_e - 1/(f_o+f_e) = \frac{f_o}{f_e(f_o+f_e)}$. $v = \frac{f_e(f_o+f_e)}{f_o}$. Magnification of eyepiece for this image $m = v/u = \frac{f_e(f_o+f_e)}{f_o(f_o+f_e)} = f_e/f_o$. Diameter of exit pupil $D_{exit} = |m| \times D_{obj} = (f_e/f_o) D_{obj}$. Hence $D_{obj} / D_{exit} = f_o / f_e = M$.
Solution: Achromatic doublet combines a convex crown lens and a concave flint lens. The flint lens has higher dispersion but is given a lower power, so it perfectly cancels the chromatic dispersion of the crown lens without completely canceling its focal power. Parabolic mirrors focus all incoming parallel rays, whether paraxial or marginal, precisely to a single geometrical point, completely eliminating spherical aberration which plagues spherical mirrors.
