CBSE · Class 12 · Physics · Ray Optics and Optical Instrumentsएक संयुक्त सूक्ष्मदर्शी (compound microscope) का नामांकित किरण आरेख खींचिए जो स्पष्ट दृष्टि की न्यूनतम दूरी पर प्रतिबिंब के बनने को दर्शाता है। इसकी आवर्धन क्षमता (magnifying power) के लिए व्यंजक की व्युत्पत्ति कीजिए।

Introduction to Compound Microscope\nA compound microscope is an optical instrument used to observe highly magnified images of tiny objects. It consists of two convex lenses: a small objective lens facing the object and an eye-piece lens of larger focal length facing the eye.

Working Principle and Ray Diagram Description

• Objective Lens ($L_o$): It has a very short focal length ($f_o$) and small aperture. It forms a real, inverted, and magnified image ($A'B'$) of the object ($AB$) at a position within the focal length of the eye-piece.
• Eye-Piece ($L_e$): It acts as a simple magnifying glass. It forms a virtual, highly magnified, and inverted final image ($A''B''$) at the least distance of distinct vision ($D$).

Derivation of Magnifying Power\nMagnifying power ($m$) is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye when both are placed at the least distance of distinct vision:

$$m = \frac{\beta}{\alpha}$| \nSince the angles are small, we can approximate $\alpha \approx \tan \alpha$ and $\beta \approx \tan \beta$: $$\tan \alpha \approx \frac{AB}{D}$$ $$\tan \beta \approx \frac{A'B'}{u_e}$| \nTherefore, magnifying power becomes: $$m = \frac{A'B' / u_e}{AB / D} = \left(\frac{A'B'}{AB}\right) \left(\frac{D}{u_e}\right)$| \nHere, $\frac{A'B'}{AB} = m_o$ is the linear magnification produced by the objective lens, given by: $$m_o = \frac{v_o}{u_o}$| \nThus: $$m = \frac{v_o}{u_o} \left(\frac{D}{u_e}\right)$| \nUsing the lens formula for the eye-piece: $$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$| \nSetting $v_e = -D$: $$-\frac{1}{D} - \frac{1}{u_e} = \frac{1}{f_e}$|\nMultiplying throughout by $D$: $$-1 - \frac{D}{u_e} = \frac{D}{f_e} \implies \frac{D}{u_e} = 1 + \frac{D}{f_e}$| \nSubstituting this into the magnification equation: $$m = \frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right)$| \nSince the object is placed very close to the focus of the objective ($u_o \approx f_o$) and the image is formed close to the eye-piece ($v_o \approx L$, where $L$ is the tube length): $$m \approx -\frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$|\nThis is the required expression for the magnifying power of a compound microscope.