Question

If the magnification of a compound microscope when the final image forms at the near point is \(14\%\) greater than its magnification when the final image forms at infinity, then the ratio of the focal length of the eyepiece to the least distance of distinct vision is

• A. \(3:25\)
• B. \(3:50\)
• C. \(7:50\)
• D. \(7:25\)

Hint:

For a compound microscope, \[ \boxed{ M_\infty=\frac{L}{f_o}\cdot\frac{D}{f_e}, \qquad M_D=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right) } \]

Answer 1

The Correct Option is C

Step 1: Write the magnification expressions. For final image at infinity, \[ M_\infty = \frac{L}{f_o}\cdot\frac{D}{f_e}. \] For final image at the near point, \[ M_D = \frac{L}{f_o} \left(1+\frac{D}{f_e}\right). \]

Step 2: Use the given condition. The magnification at the near point is \(14\%\) greater than that at infinity. \[ M_D = 1.14\,M_\infty. \] Substituting, \[ 1+\frac{D}{f_e} = 1.14\left(\frac{D}{f_e}\right). \] Let \[ x=\frac{D}{f_e}. \] Then, \[ 1+x=1.14x, \] \[ 1=0.14x, \] \[ x=\frac{50}{7}. \] Hence, \[ \frac{f_e}{D} = \frac{7}{50}. \] Therefore, \[ \boxed{f_e:D=7:50.} \] Thus, \[ \boxed{(C)} \] is the correct answer.