Lens Makers Formula/ Len's Formula: Practice Problem & Solution
In a compound microscope, the focal length of two lenses are \(1.25 \text{cm}\) and \(12.5 \text{cm}\). If an object is placed at \(5 \text{cm}\) from objective lens and final image is formed at \(25 \text{cm}\) from eye piece lens, the distance between the two lenses is
Solution Explained:
To solve this problem, we apply the core principles of Lens Makers Formula/ Len's Formula. Understanding the underlying formula is key to arriving at the correct answer below:
$$\begin{aligned} &\text{For the objective lens: } \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \\ &\frac{1}{v_o} - \frac{1}{-5} = \frac{1}{1.25} \implies v_o = 1.67 \text{ cm} \\ &\text{For the eyepiece lens: } \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \\ &\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{12.5} \implies u_e = -8.33 \text{ cm} \\ &\text{Total distance } L = \vert{}v_o\vert{} + \vert{}u_e\vert{} = 1.67 + 8.33 = 10 \text{ cm} \end{aligned}$$
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