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: Understanding the Question:
We need to calculate the magnifying power (M) of a compound microscope under normal adjustment (which means the final image is formed at infinity). Then, we equate the result to $5^\alpha$ to find the exponent $\alpha$.
Step 2: Key Formula or Approach:
For a compound microscope in normal adjustment (image at infinity), the magnifying power is given by the formula:
$M = \frac{L}{f_{o}} \times \frac{D}{f_{e}}$
where:
$L$ = length of the microscope tube
$f_{o}$ = focal length of the objective lens
$f_{e}$ = focal length of the eyepiece lens
$D$ = least distance of distinct vision (standard value is 25 cm)
Step 3: Detailed Explanation:
Given values:
Focal length of objective, $f_{o} = 2 \text{ cm}$ (smaller focal length is objective)
Focal length of eyepiece, $f_{e} = 5 \text{ cm}$
Tube length, $L = 10 \text{ cm}$
Least distance of distinct vision, $D = 25 \text{ cm}$
Substitute these values into the magnifying power formula:
$M = \frac{L}{f_{o}} \times \frac{D}{f_{e}}$
$M = \frac{10}{2} \times \frac{25}{5}$
$M = 5 \times 5 = 25$
The question states that the magnifying power is $5^\alpha$.
So, $5^\alpha = 25$
$5^\alpha = 5^2$
Comparing the exponents, we get:
$\alpha = 2$
Step 4: Final Answer:
The value of $\alpha$ is 2.