Calculate the magnification of this telescope.

Outline why sign convention is necessary in optics.

A student decides to reverse the positions of the same lenses without changing the separation to form an optical microscope in normal adjustment. The student’s near point is 0.25 m from her eye.

(i) Show, using a calculation, that the image formed by the objective lens is about 0.19 m from the eyepiece.

(ii) Calculate the distance between the objective lens of the microscope and the object.

(iii) Determine the overall magnification of the microscope.

F_o+f_e=84 so f_e=84-82=2cm

$\ll M = \frac{{{f_{\rm{O}}}}}{{{f_{\rm{e}}}}} = \frac{{82}}{2} =  \gg 41$

a sign convention is a way to distinguish between real and virtual objects or images or converging and diverging lenses

(i) image will be virtual v=–25 cm

$\frac{1}{u} = \frac{1}{{82}} + \frac{1}{{25}}$

«=19cm or 0.19m»

Award [1 max] if v = +25 cm used to give u = –36 cm.

(ii) image will be real v=84-19=65cm
$\ll \frac{1}{u} = \frac{1}{2} - \frac{1}{{65}} \gg$ so u=2.1cm

(iii) ${M_e} =  \ll \frac{D}{{{f_{\rm{e}}}}} + 1 = \frac{{25}}{{82}} + 1 =  \gg 1.3$ AND ${m_o} =  \ll \frac{v}{{{f_{\rm{o}}}}} - 1 = \frac{{65}}{2} - 1 =  \gg 31$ or 32

so $M =  \ll {M_{\rm{e}}}{m_{\rm{o}}} = 1.3 \times 31 =  \gg 40$ or 41

Far point adjustment gives M = 9.3 (accept answers from interval 9.3 to 9.6), award [1 max] for full working.