A compound microscope has an objective and an eyepiece of focal lengths $ f_0 $ and $ f_e $, respectively. To obtain a large magnification of a small object, the microscope should have:

Question

Object is placed at $40 \text{ cm}$ from spherical surface whose radius of curvature is $20 \text{ cm}$. Find height of image formed.

Answer 1

The magnification $ M $ of a compound microscope is calculated as the product of the angular magnification of the eyepiece and the magnification of the objective. This can be expressed as: \[M = \frac{\text{angular magnification of the eyepiece} \times \text{magnification of the objective}}{1}\] The magnification of the objective lens is determined by the formula: \[M_o = \frac{D}{f_0}\] Here, $ D $ represents the least distance of distinct vision, typically 25 cm, and $ f_0 $ is the focal length of the objective lens. The magnification of the eyepiece is calculated using: \[M_e = \frac{D}{f_e}\] In this equation, $ f_e $ is the focal length of the eyepiece. To achieve high magnification, both $ f_0 $ and $ f_e $ must be small. A smaller $ f_0 $ results in greater magnification by the objective, and a slightly larger $ f_e $ (where $ f_e>f_0 $) is necessary for the eyepiece to effectively magnify the image produced by the objective. Consequently, a compound microscope requires both small focal lengths for the objective ($ f_0 $) and the eyepiece ($ f_e $), with $ f_e $ being greater than $ f_0 $, to attain significant magnification. Thus, the correct answer is option (A).

A compound microscope has an objective and an eyepiece of focal lengths \( f_0 \) and \( f_e \), respectively. To obtain a large magnification of a small object, the microscope should have:

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The Correct Option is A

Solution and Explanation

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