Question

Draw a labelled ray diagram of a compound microscope showing image formation at least distance of distinct vision. Derive an expression for its magnifying power.

Hint:

For a compound microscope, the magnifying power is the product of the magnifying powers of the objective and the eyepiece. The objective magnifies the object, and the eyepiece further magnifies the image formed.

Answer 1

In a compound microscope, the objective lens forms a real, inverted, and diminished image at the focal plane of the eyepiece. The eyepiece acts as a magnifier to form a virtual, erect, and magnified image at the least distance of distinct vision. The magnifying power \( M \) of the compound microscope is given by the product of the magnifying powers of the objective lens \( M_o \) and the eyepiece lens \( M_e \): \[ M = M_o \times M_e. \] The magnifying power of the objective lens is given by: \[ M_o = \frac{v_o}{u_o}, \] where \( v_o \) is the image distance and \( u_o \) is the object distance for the objective lens. Since the image is formed at the focal length of the objective lens \( f_o \), we have: \[ v_o = f_o. \] For the eyepiece, the magnifying power is given by: \[ M_e = \frac{D}{f_e}, \] where \( D \) is the least distance of distinct vision and \( f_e \) is the focal length of the eyepiece. Thus, the total magnifying power is: \[ M = \frac{f_o}{u_o} \times \frac{D}{f_e}. \]