Focal length of objective lens and eyepiece lens are 1.25 cm and 5 cm and tube length is 26 cm. Find magnification of compound microscope in normal adjustment.

The correct answer is 104.

To find the magnification of a compound microscope in normal adjustment (when the final image is formed at infinity), we can use the magnifying power formula:
$m=\frac{L}{f_o}\times \frac{D}{f_e}$

Where:
• $L$ is the tube length of the microscope = 26 cm
• $f_o$ is the focal length of the objective lens = 1.25 cm
• $f_e$ is the focal length of the eyepiece lens = 5 cm
• $D$ is the least distance of distinct vision = 25 cm

Substituting the given values into the formula:
$m=\frac{26}{1.25}\times \frac{25}{5}$

We can simplify the equation step-by-step:
First, calculate the objective lens magnification term:
$\frac{26}{1.25}=20.8$
Next, calculate the eyepiece lens magnification term:
$\frac{25}{5}=5$

Now, multiply the two parts to get the overall magnification:
$m=20.8\times 5=104$

Thus, the magnification of the compound microscope in normal adjustment is 104.