Question

Draw a labelled ray diagram showing the formation of image by a compound microscope when final image is formed at least distance of distinct vision. Derive an expression for its magnifying power for this case.

Hint:

Describe the ray diagram with the real first image and the virtual final image at D. Multiply the objective magnification by the eyepiece magnification, 1 + D/f_e.

Answer 1

Step 1: Understanding the Concept:
A compound microscope has two convex lenses. The objective lens has a very short focal length \(f_o\) and the eyepiece has a larger focal length \(f_e\).
The small object is placed just outside the focus of the objective. The objective forms a real, inverted and enlarged first image.
The eyepiece works like a simple magnifier on this image. For the final image at the least distance of distinct vision \(D\), the first image lies inside the focal length of the eyepiece.

Step 2: Ray diagram (described in words):
1. Draw the principal axis. Place the objective \(L_o\) on the left and the eyepiece \(L_e\) on the right, separated by a distance (tube length).
2. Mark the object AB, a small vertical arrow, just beyond \(F_o\) on the left of the objective.
3. Draw a ray from B parallel to the axis. It passes through the focus \(F_o'\) on the other side of the objective. Draw a second ray from B through the optical centre, going straight.
4. These two rays meet at B', giving the real, inverted and enlarged image A'B' between \(F_e\) and the eyepiece.
5. Draw rays from B' through the eyepiece. They diverge after refraction. Extend them backwards to meet at B'', which is the final virtual, enlarged image A''B''.
6. Label the final image as lying at distance \(D\) (25 cm) from the eye, on the same side as the object. It is inverted with respect to the object.

Step 3: Derivation of the magnifying power:
Magnifying power is \(m = m_o \times m_e\), where \(m_o\) is the lateral magnification of the objective and \(m_e\) is that of the eyepiece.
Objective: \(m_o = \dfrac{h'}{h} = \dfrac{v_o}{u_o}\) in magnitude, where \(h\) is the object size and \(h'\) is the size of the first image.
Eyepiece: the final image is at \(v_e = -D\). The lens formula gives \(\dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}\).
\[ \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{D} - \frac{1}{f_e} \quad\Rightarrow\quad \frac{D}{|u_e|} = 1 + \frac{D}{f_e} \]
Eyepiece magnification: \(m_e = \dfrac{v_e}{u_e} = \dfrac{D}{|u_e|} = 1 + \dfrac{D}{f_e}\).
\[ m = m_o\,m_e = \frac{v_o}{|u_o|}\left(1 + \frac{D}{f_e}\right) \]
For a microscope \(|u_o| \approx f_o\) and \(v_o \approx L\), the distance between the lenses (tube length). So:
\[ m \approx \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \]

Final Answer:
For the final image at the near point D, the magnifying power is the product of the two magnifications. The result is: \[ \boxed{m = \frac{v_o}{|u_o|}\left(1 + \frac{D}{f_e}\right) \approx \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)} \]