Question

The eyepiece of a microscope has a focal length of \(1.25 \mathrm{cm}\) and the objective lens focal length is \(1.44 \mathrm{cm} .\) (a) If the tube length is \(18.0 \mathrm{cm},\) what is the angular magnification of the microscope? (b) What objective focal length would be required to double this magnification?

Short answer

Answer: The angular magnification of the microscope is -11.5. Question: What would be the required focal length of the objective lens in order to double the microscope's magnification? Answer: A focal length of approximately 0.61 cm for the objective lens is required to double the magnification of the microscope.

Step-by-step solution

Understand angular magnification formula for a microscope

The angular magnification (M) of a microscope can be found using the formula: \( M = - \dfrac{f_e (L - f_o)}{f_o f_e} \) where \(f_e\) is the focal length of the eyepiece, \(f_o\) is the focal length of the objective lens, and \(L\) is the tube length.

Calculate the angular magnification for given values

Now, we plug in the values given in the exercise for \(f_e = 1.25 \mathrm{cm}\), \(f_o = 1.44 \mathrm{cm}\), and \(L = 18.0 \mathrm{cm}\). \( M = - \dfrac{1.25 (18 - 1.44)}{1.44 \cdot 1.25} \) Now, perform the arithmetic operations to get the value of \(M\): \( M = - \dfrac{1.25 (16.56)}{1.44 \cdot 1.25} = - \dfrac{20.7}{1.8} \approx -11.5 \) So, the angular magnification of the microscope is -11.5. (Note that the negative sign indicates that the image is inverted).

Calculate the required objective focal length to double the magnification

We're asked to find the objective focal length required to double the magnification, which means we want a magnification of \(-2 \times M = -2 \times(-11.5) = 23\). Let's denote the new objective focal length as \(f'_o\). So, \( 23 \approx - \dfrac{1.25(18 - f'_o)}{f'_o \cdot 1.25} \) Now, we need to solve for \(f'_o\). First, we simplify and multiply both sides by \(-1.25 f'_o\): \( 23 \cdot -1.25 f'_o = -(18 - f'_o) \) Continue solving for \(f'_o\): \( -28.75 f'_o = -18 + f'_o \) \( -29.75 f'_o = -18 \) \( f'_o \approx \dfrac{18}{29.75} = 0.61 \mathrm{cm} \) Therefore, an objective lens focal length of approximately \(0.61 \mathrm{cm}\) would be required to double the magnification of the microscope.