A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.

### Related Formula textMagnification m = fracD_textoutD_textin = fracf_2f_1$$\text{Magnification } m = \frac{D_{\text{out}}}{D_{\text{in}}} = \frac{f_2}{f_1}$$ ### Core Logic For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.

By similar triangles at the focal point: tan theta = fracD_1 / 2f_1 = fracD_2 / 2f_2$$\tan \theta = \frac{D_1 / 2}{f_1} = \frac{D_2 / 2}{f_2}$$ ### Step 1: Solving for f2 fracD_1f_1 = fracD_2f_2$$\frac{D_1}{f_1} = \frac{D_2}{f_2}$$ frac240 = frac14f_2$$\frac{2}{40} = \frac{14}{f_2}$$ f_2 = 14 times frac402 = 14 times 20 = 280text mm$$f_2 = 14 \times \frac{40}{2} = 14 \times 20 = 280\text{ mm}$$ ### Pattern Recognition A standard beam expander (Keplerian telescope design used in reverse) has f_2/f_1 = D_2/D_1$f_2/f_1 = D_2/D_1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments