DP Physics (last assessment 2024)

Question 19N.3.SL.TZ0.8

[Maximum mark: 6]

19N.3.SL.TZ0.8

A small object is placed at a distance of 2.0 cm from the objective lens of an optical compound microscope in normal adjustment.

The following data are available.

Magnification of the microscope = 70
Focal length of the eyepiece = 3.0 cm
Near point distance = 24 cm

(a)

State what is meant by normal adjustment when applied to a compound microscope.

[1]

Markscheme

«the final» image is formed at the near point of the eye ✔

(b)

Calculate, in cm, the distance between the eyepiece and the image formed by the objective lens.

[2]

Markscheme

«image is virtual so» $v = - 24$ «cm» ✔

« $\frac{1}{u} = \frac{1}{3.0} + \frac{1}{24}$ » so $u = 27$ «cm» ✔

(c)

Determine, in cm, the focal length of the objective lens.

[3]

Markscheme

$M_{e} = \frac{v}{u} = \frac{24}{2.66} = 9.0$ AND $M_{o} = \frac{7.0}{9.0} = 7.8$ ✔

$v_{0} = 2.0 \times 7.8 = 15.6 « cm »$ ✔

$« \frac{1}{f} = \frac{1}{2} + \frac{1}{16} »$ so $f_{0} = 1.8 « cm »$ ✔

NOTE: MP1 allow $M_{e} = \frac{D}{f} + 1 = 9$

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