SEMI FINAL STAGE 2025
St. Augustine's College: 46 points
Pope John Seminary SHS: 38 points
Amaniampong SHS: 23 points
ROUND 2 - SPEED RACE
FIRST QUESTION
A compound microscope has an objective lens of focal length 0.50 cm and an eyepiece of focal length 2.5 cm, with the two lenses separated by 20.0 cm.
The object is placed 0.52 cm from the objective, and the final image is formed at infinity.
Find the overall angular magnification.
ANSWER: $M = -250$
FORMULA
$M = -\dfrac{f_{o}}{d_{o} - f_{o}} \times \dfrac{25}{f_{e}}$
SOLUTION:
$M = -\dfrac{0.50}{0.52 - 0.50} \times \dfrac{25}{2.5}$
$M = -\dfrac{0.50}{0.02} \times 10$
$M = -25 \times 10$
$M = -250$
ANSWER: $M = -250$
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PRACTICE QUESTIONS
1. QUESTION: A compound microscope has an objective of focal length 0.60 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.64 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −150 or $-1.5\times10^{2}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.64 - 0.60$
$u - f_o = 0.04\text{ cm}$
$M = -\dfrac{0.60 \times 25}{0.04 \times 2.5}$
$M = -\dfrac{6\times10^{-1} \times 25}{4\times10^{-2} \times 25\times10^{-1}}$
$M = -\dfrac{3}{2} \times10^{2}$
$M = -1.5\times10^{2}$
2. QUESTION: A compound microscope has an objective of focal length 0.60 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.65 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −120 or $-1.2\times10^{2}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.65 - 0.60$
$u - f_o = 0.05\text{ cm}$
$M = -\dfrac{0.60 \times 25}{0.05 \times 2.5}$
$M = -\dfrac{6\times10^{-1} \times 25}{5\times10^{-2} \times 25\times10^{-1}}$
$M = -\dfrac{6}{5} \times10^{2}$
$M = -1.2\times10^{2}$
3. QUESTION: A compound microscope has an objective of focal length 0.70 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.75 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −140 or $-1.4\times10^{2}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.75 - 0.70$
$u - f_o = 0.05\text{ cm}$
$M = -\dfrac{0.70 \times 25}{0.05 \times 2.5}$
$M = -\dfrac{7\times10^{-1} \times 25}{5\times10^{-2} \times 25\times10^{-1}}$
$M = -\dfrac{7}{5} \times10^{2}$
$M = -1.4\times10^{2}$
4. QUESTION: A compound microscope has an objective of focal length 0.30 cm and an eyepiece of focal length 5.0 cm. The object is placed 0.32 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −75 or $-7.5\times10^{1}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.32 - 0.30$
$u - f_o = 0.02\text{ cm}$
$M = -\dfrac{0.30 \times 25}{0.02 \times 5.0}$
$M = -\dfrac{3\times10^{-1} \times 25}{2\times10^{-2} \times 5}$
$M = -\dfrac{3 \times 5}{2} \times10^{1}$
$M = -75$
5. QUESTION: A compound microscope has an objective of focal length 0.45 cm and an eyepiece of focal length 5.0 cm. The object is placed 0.50 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −45 or $-4.5\times10^{1}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.50 - 0.45$
$u - f_o = 0.05\text{ cm}$
$M = -\dfrac{0.45 \times 25}{0.05 \times 5.0}$
$M = -\dfrac{45 \times 25}{5 \times 5}$
$M = -9 \times 5$
$M = -45$
6. QUESTION: A compound microscope has an objective of focal length 0.44 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.48 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −110 or $-1.1\times10^{2}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.48 - 0.44$
$u - f_o = 0.04\text{ cm}$
$M = -\dfrac{0.44 \times 25}{0.04 \times 2.5}$
$M = -\dfrac{44\times10^{-2} \times 25}{4\times10^{-2} \times 25\times10^{-1}}$
$M = -1.1\times10^{2}$
7. QUESTION: A compound microscope has an objective of focal length 0.55 cm and an eyepiece of focal length 5.0 cm. The object is placed 0.60 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −55 or $-5.5\times10^{1}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.60 - 0.55$
$u - f_o = 0.05\text{ cm}$
$M = -\dfrac{0.55 \times 25}{0.05 \times 5.0}$
$M = -\dfrac{55 \times 25}{5 \times 5}$
$M = -11 \times 5$
$M = -55$
8. QUESTION: A compound microscope has an objective of focal length 0.65 cm and an eyepiece of focal length 2.5 cm. The object is placed 0.70 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −130 or $-1.3\times10^{2}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.70 - 0.65$
$u - f_o = 0.05\text{ cm}$
$M = -\dfrac{0.65 \times 25}{0.05 \times 2.5}$
$M = -\dfrac{65\times10^{-2} \times 25}{5\times10^{-2} \times 25\times10^{-1}}$
$M = -1.3\times10^{2}$
9. QUESTION: A compound microscope has an objective of focal length 0.21 cm and an eyepiece of focal length 5.0 cm. The object is placed 0.24 cm from the objective and the final image is formed at infinity. Find the overall angular magnification.
ANSWER: −35 or $-3.5\times10^{1}$
SOLUTION:
$M = -\dfrac{f_o}{u - f_o} \times \dfrac{25}{f_e}$ (final image inverted)
$u - f_o = 0.24 - 0.21$
$u - f_o = 0.03\text{ cm}$
$M = -\dfrac{0.21 \times 25}{0.03 \times 5.0}$
$M = -\dfrac{21 \times 25}{3 \times 5}$
$M = -7 \times 5$
$M = -35$