2020 A/L Physics Past Paper · Question 36 (English Medium)

Answer: (1) $4$

Given information

We know:

Total magnifying power in normal adjustment $M = 100$.

Focal length of the objective $f_o = 2.5$ cm; object distance $u_1 = 2.6$ cm.

$D = 25$ cm.

What the question asks

We want: the eyepiece's magnification $\left(1 + \dfrac{D}{f_e}\right)$ — given the total magnifying power ($100$) and the objective's data.

Relevant theory

The compound microscope uses two lenses to magnify the image twice to obtain a higher magnification.

When the eye-piece is adjusted to have the final image at the near point, the microscope is in normal adjustment.

In normal adjustment the final image forms at the near point ($D = 25$ cm).

Using 1/v − 1/u = 1/f and applying the sign convention, 1/v − 1/u = 1/(−f)

Required equations

$\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f_o}$

$m_o = \frac{v_1}{u_1}$

$M = \left(1 + \frac{D}{f_e}\right) m_o$

Applying it to this question

Plan:

Find $v_1$ for the objective from the lens equation.

Find $m_o = v_1/u_1$.

Find the eyepiece's magnification from $M = (1 + D/f_e)\,m_o$.

Reasoning and calculation

$\frac{1}{v_1} = \frac{1}{2.5} - \frac{1}{2.6} = \frac{0.1}{6.5}$

$65$

$v_1 = 65$

$25$

$m_o = \frac{65}{2.6}$

$4$

$1 + \frac{D}{f_e} = \frac{M}{m_o} = \frac{100}{25}$

Checks

Unit check: $v_1/u_1$ $has\,cm/cm$ — magnification is dimensionless ✓; $100 = 25 \times 4$ ✓.

Does it make sense? The object is only $0.1$ cm outside the focus — so the objective's image is far away ($65$ cm) and large ($\times 25$). The eyepiece then acts as a simple magnifier of $4$ times — $f_e \approx 8.3$ cm, a reasonable size ✓.

$100 * (u - f) / f$

A second route: $m_o = \dfrac{v_1}{u_1} = \dfrac{f_o}{u_1 - f_o} = \dfrac{2.5}{0.1} = 25$ — the same value; eyepiece is $4$. (From this $f_e = \dfrac{25}{4 - 1} \approx 8.3$ cm, $u_2 = \dfrac{25}{4} = 6.25$ cm — a sensible microscope.)

Conclusion

$1 + \frac{D}{f_e} = \frac{100}{25} = 4$

The objective's magnification is $25$; so the eyepiece's magnification is $100 \div 25 = 4$ — the answer is (1).

Taking the objective's magnification ($25$) as the answer.

$Total = objective \times eyepiece$; eyepiece $= 100 / 25 = 4$.