Question

Draw the ray diagram of a compound microscope. Find the expression for its magnifying power.

Hint:

Magnifying power is the product of objective magnification \( L/f_o \) and eyepiece magnification \( (1 + D/f_e) \).

Answer 1

Step 1: Ray diagram (description).
A compound microscope uses two convex lenses: an objective of small focal length \( f_o \) and an eyepiece of slightly larger focal length \( f_e \), fixed at the two ends of a tube. The small object \( AB \) is placed just beyond the focus of the objective. The objective forms a real, inverted and magnified image \( A'B' \) inside the tube, between the eyepiece and its focus. This image \( A'B' \) acts as the object for the eyepiece, which works like a simple microscope and forms a virtual, inverted and highly magnified final image \( A''B'' \) at the least distance of distinct vision \( D \) (about 25 cm) from the eye.
(Ray diagram: object AB near objective focus, refracted rays meet to form real image A'B' inside tube, then rays from A'B' pass through the eyepiece and diverge, their backward extensions locating the enlarged virtual image A''B'' on the same side as the object.)

Step 2: Magnifying power definition.
Magnifying power \( M \) is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object when placed at distance \( D \). It equals the product of the linear magnification of the objective \( m_o \) and the angular magnification of the eyepiece \( m_e \):
\[ M = m_o \times m_e \]

Step 3: Magnification by the objective.
\[ m_o = \frac{v_o}{u_o} = \frac{L}{f_o} \]
where \( v_o \) and \( u_o \) are the image and object distances for the objective and \( L \) is the tube length (the distance between the objective image and the objective, taken nearly equal to the tube length).

Step 4: Magnification by the eyepiece.
The eyepiece acts as a simple microscope with the final image at the near point \( D \):
\[ m_e = 1 + \frac{D}{f_e} \]

Step 5: Total magnifying power.
\[ M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \]
When the final image is formed at infinity (relaxed eye), \( m_e = D/f_e \) and
\[ M = \frac{L}{f_o}\cdot\frac{D}{f_e} \]

\[\boxed{\ M = \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right)\ \text{(image at near point)}\ }\]