Past exam of the physics 2 paper of the University of Oxford / 2021 / 4

Use positive real object and image distances $u,v$ for a thin lens. The required absolute magnification is $M=1000/15=200/3$, and the specified LCD-to-screen separation gives $u+v=2000\mathrm{mm}$, not $v=2000\mathrm{mm}$. Since $v/u=M$,The thin lens equation yieldsPlace the projection lens $29.56\mathrm{mm}$ beyond the LCD towards the screen. Its real image is inverted; the signed magnification is $-200/3$.

The condenser lens should image the point source onto the small projection lens. Then every illuminated LCD point directs its central light ray into the projection lens, rather than wasting light elsewhere. In the paraxial idealization this gives approximately uniform useful illumination.

Put the LCD at $z=0$, the source at $z=-100\mathrm{mm}$, the projection lens at $z=u=29.5567\mathrm{mm}$, and the condenser lens at $z=-h$. The cone from the condenser lens's rim, of radius $15\mathrm{mm}$, to the projection lens has radiusat the LCD. Maximum approximately uniform illumination without unnecessary overfilling makes this equal the half-diagonal $7.5\mathrm{mm}$. Thus $h=u$: place the condenser lens $29.56\mathrm{mm}$ before the LCD, or $70.44\mathrm{mm}$ after the source. Its object and image distances are $100-h$ and $u+h$, soThe illuminated disk just contains the rectangular LCD. Exact spatially uniform irradiance is not guaranteed by a single spherical lens and a point source; the result uses paraxial geometrical optics, negligible optical aberration, and an approximately uniform incident cone.