A converging lens is used as a magnifying glass. On the diagram draw rays to construct the image of the object, o.

The lens has a focal length f. When the image is formed at the near point, the distance u of the object from the lens is given by

\[u = \frac{{fD}}{{D + f}}\]

where D is the near point distance.

Deduce that the angular magnification M is given by

\[M = 1 + \frac{D}{f}\]

A compound microscope consists of an eyepiece lens of focal length 6.0 cm and an objective lens of focal length 2.8 cm. An object is placed 3.4 cm from the objective lens and the final image of the object is formed by the microscope at the near point.

Determine the

(i) angular magnification of the eyepiece. Take the near point distance to be 25 cm.

(ii) distance from the objective lens of the intermediate image formed by this lens.

(iii) overall magnification of the compound microscope.

ray 1 correct;
ray 2 correct;
virtual rays converge/image shown;
Ignore arrows on any lines drawn.

\(M = \left( {\frac{{{\theta _{\rm{i}}}}}{{{\theta _{\rm{o}}}}} = } \right)\left(  -  \right)\frac{v}{u}\);
\( = \frac{D}{{\frac{{fD}}{{f + D}}}} = \frac{{D\left( {f + D} \right)}}{{fD}}\);
\(\left( {M = 1 + \frac{D}{f}} \right)\)
Check for correct manipulation.

(i) \(M = \left( {1 + \frac{{25}}{6} = } \right)5.2\);

(ii) \(\frac{1}{v} = \frac{1}{{2.8}} - \frac{1}{{3.4}}\);
v=16 cm;
Award [2] for a bald correct answer.
Award [1] for ECF giving v =1.5 cm.

(iii) magnification of objective\( = \left( {\frac{{16}}{{3.4}} = } \right)4.7\);
overall magnification = (5.2×4.7 =)24;
Award [2] for a bald correct answer.
Award [2 max] for ECF from (d)(i) and (d)(ii).