Question

There is a compound microscope of lenses having focal lengths 2 cm and 5 cm and tube length 10 cm. Find magnifying power in normal adjustment. If your answer is $5^\alpha$, find '$\alpha$' :

Hint:

"Normal adjustment" always implies relaxed eye viewing, which means the final image is at infinity. The term "tube length" (L) in the standard formula $M = (L/f_o) \times (D/f_e)$ strictly represents the distance between the secondary focal point of the objective and the primary focal point of the eyepiece, but in textbook problems, it's often approximated as the distance between the two lenses.

Answer 1

Correct Answer: 2

Step 1: Break magnifying power into objective and eyepiece parts

For a compound microscope under normal adjustment, the total magnifying power is the product of:

• Linear magnification of the objective
• Angular magnification of the eyepiece

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Step 2: Magnification due to objective

The linear magnification produced by the objective is approximately:

m_o = L / f_o

Given:

L = 10 cm
f_o = 2 cm

m_o = 10 / 2 = 5

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Step 3: Magnification due to eyepiece

For normal adjustment, the angular magnification of the eyepiece is:

m_e = D / f_e

Given:

D = 25 cm
f_e = 5 cm

m_e = 25 / 5 = 5

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Step 4: Calculate total magnifying power

M = m_o × m_e

M = 5 × 5 = 25

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Step 5: Express magnifying power in the form 5^α

25 = 5²

So,

α = 2

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Final Answer:

The value of α is
2