In a microscope the objective is having focal length \(f_0=2 cm\) and eye-piece is having focal length \(f_{ e }=4 cm\). The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is _________ .

Explanation:
Let us clarify the concept of magnification under different adjustments step-by-step based on the Ray Optics and Optical Instruments chapter of the NCERT Class 12 Physics textbook:

Step 1: Understand the meaning of "Normal Adjustment"
In optical instruments, "Normal Adjustment" explicitly means that the final image is formed at infinity \(\infty\), allowing the observer's eye to view it in a completely relaxed state.

Step 2: Apply the correct formula for Normal Adjustment
When the final image is formed at infinity, the magnifying power \(m\) of a compound microscope is given by:
\(m = \frac{L}{f_o} \times \frac{D}{f_e}\)
Where:
- \(L\) = Tube length = 32 cm
- \(f_o\) = Focal length of objective = 2 cm
- \(f_e\) = Focal length of eyepiece = 4 cm
- \(D\) = Least distance of distinct vision = 25 cm

Step 3: Calculate the value
Substitute the given values into the infinity-adjustment formula:
\(m = \frac{32}{2} \times \frac{25}{4}\)
\(m = 16 \times 6.25 = 100\)

Why 116 is incorrect here:
The handwritten solution uses the formula \(m = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\). This formula belongs to the near-point adjustment, where the final image is formed at the least distance of distinct vision (\(D = 25\text{ cm}\)). Since this specific question asks for normal adjustment, using that formula is a conceptual error. The app's answer of 100 is fully correct.

Hence, Option B is the correct choice.