1. **Problem statement:** Consider a two-lens system in a 4-f configuration with lenses of focal lengths $f_1$ and $f_2$ separated by $f_1 + f_2$. An object is placed at the front focal plane of the first lens. 2. **Goal:** (a) Show the image forms at the back focal plane of the second lens and find the transverse magnification. 3. **Lens formula and 4-f system:** For a thin lens, the lens formula is: $$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$$ where $s$ is object distance, $s'$ is image distance, and $f$ is focal length. In a 4-f system, the lenses are separated by $f_1 + f_2$. 4. **Step (a): Image location and magnification** - The object is at the front focal plane of lens 1, so $s = f_1$. - Using the lens formula for lens 1: $$\frac{1}{f_1} + \frac{1}{s'_1} = \frac{1}{f_1} \implies \frac{1}{s'_1} = 0 \implies s'_1 = \infty$$ - The rays after lens 1 are collimated (parallel). - These parallel rays travel distance $f_1 + f_2$ to lens 2. - Lens 2 focuses parallel rays to its back focal plane at distance $s'_2 = f_2$. - Therefore, the final image is at the back focal plane of lens 2. 5. **Transverse magnification $M$:** - Magnification of lens 1: $$M_1 = -\frac{s'_1}{s} = -\frac{\infty}{f_1} \to \text{undefined but rays are parallel, so consider angular magnification}$$ - For the 4-f system, the overall transverse magnification is: $$M = -\frac{f_2}{f_1}$$ - The negative sign indicates image inversion. 6. **Answer for (a):** An object at the front focal plane of lens 1 produces an image at the back focal plane of lens 2 with transverse magnification: $$M = -\frac{f_2}{f_1}$$ 7. **Step (b): Speed of image $v_2$ when object moves at speed $v_1$** - The image position $x_i$ relates to object position $x_o$ by: $$x_i = M x_o$$ - Differentiating w.r.t time $t$: $$v_2 = \frac{dx_i}{dt} = M \frac{dx_o}{dt} = M v_1$$ - Substitute $M$: $$v_2 = -\frac{f_2}{f_1} v_1$$ 8. **Answer for (b):** The image speed is: $$v_2 = -\frac{f_2}{f_1} v_1$$ 9. **Step (c): Tilt angle transformation** - A ray tilted by angle $\theta_1$ entering lens 1 will be transformed by the system. - The angular magnification of a 4-f system is: $$\theta_2 = -\frac{f_1}{f_2} \theta_1$$ - The negative sign indicates the image tilt is inverted relative to the object tilt. 10. **Answer for (c):** The tilt angle of the image after the second lens is: $$\theta_2 = -\frac{f_1}{f_2} \theta_1$$