CBSE 2025, Set 2

Question

(a) (i) Draw a ray diagram to show the image formation by a compound microscope. Obtain the expression for the total magnification of the microscope when the final image is formed at infinity.
(ii) In a compound microscope, an object is placed at a distance of $1.5	ext{ cm}$ from the objective of focal length $1.25	ext{ cm}$. The eyepiece has a focal length of $5	ext{ cm}$. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Compound Microscope Ray Diagram & Magnification:**

![Compound Microscope Ray Diagram](https://www.edvaya.com/images/cbse/physics/2025/set2/q31a-compound-microscope-ray-diagram-answer.webp)

* **Objective Lens:** Forms a real, inverted, and magnified intermediate image ($A'B'$) of the object ($AB$).
 Magnification by objective, $m_o = \frac{v_o}{u_o}$. Approximately, if the object is very close to $f_o$, $m_o \approx \frac{L}{f_o}$ where $L$ is the tube length.
* **Eyepiece:** Acts as a simple magnifier. For the final image to form at infinity, the intermediate image $A'B'$ must fall precisely on the principal focus ($f_e$) of the eyepiece.
 Magnification by eyepiece for image at infinity, $m_e = \frac{D}{f_e}$.

The total magnification ($M$) is the product of the magnifications:
$$M = m_o 	imes m_e$$
$$\boxed{M = \left(\frac{L}{f_o}ight)\left(\frac{D}{f_e}ight)}$$

**(ii) Calculation of Lens Separation:**
Given:
* $u_o = -1.5	ext{ cm}$ (object distance for objective)
* $f_o = +1.25	ext{ cm}$
* $f_e = +5	ext{ cm}$

For the objective lens, using the thin lens formula:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$
$$\frac{1}{v_o} - \frac{1}{-1.5} = \frac{1}{1.25}$$
$$\frac{1}{v_o} + \frac{1}{1.5} = \frac{1}{1.25}$$
$$\frac{1}{v_o} = \frac{1}{1.25} - \frac{1}{1.5}$$
Using fractions: $1.25 = \frac{5}{4} \Rightarrow \frac{1}{1.25} = \frac{4}{5}$, and $1.5 = \frac{3}{2} \Rightarrow \frac{1}{1.5} = \frac{2}{3}$.
$$\frac{1}{v_o} = \frac{4}{5} - \frac{2}{3} = \frac{12 - 10}{15} = \frac{2}{15}$$
$$v_o = \frac{15}{2} = 7.5	ext{ cm}$$

Since the final image is formed at infinity, the intermediate image must act as an object located exactly at the focal point of the eyepiece. 
Thus, the object distance for the eyepiece is $|u_e| = f_e = 5	ext{ cm}$.

The distance between the two lenses ($L$) is the sum of the image distance of the objective and the object distance of the eyepiece:
$$L = |v_o| + |u_e|$$
$$L = 7.5	ext{ cm} + 5	ext{ cm}$$
$$\boxed{L = 12.5	ext{ cm}}$$

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