CBSE · Class 12 · Physics · Ray Optics and Optical InstrumentsExplain the working principle, construction, and magnifying power of a compound microscope with the help of a neat ray diagram. Derive the expression for its magnifying power when the final image is formed at the least distance of distinct vision.

Introduction\nA compound microscope is an optical instrument used to see highly magnified images of tiny objects. It consists of two convex lenses of short focal lengths: the objective lens and the eyepiece.

Construction

1. Objective Lens: A convex lens of very small focal length ($f_o$) and small aperture, facing the object.
2. Eyepiece: A convex lens of slightly larger focal length ($f_e$) and larger aperture, through which the final image is viewed.
3. Tube: Both lenses are fitted at the ends of a sliding metal tube at a suitable distance apart.

Working Principle

• The objective lens forms a real, inverted, and magnified image of the tiny object placed just beyond its focus.
• This real image acts as a virtual object for the eyepiece, which produces a virtual, highly magnified, and inverted final image with respect to the original object.

Derivation of Magnifying Power\nMagnifying power ($m$) is defined as the ratio of the angle $\beta$ subtended by the final image at the eye to the angle $\alpha$ subtended by the object when placed at the least distance of distinct vision ($D$).

$$m = \frac{\beta}{\alpha} \approx \frac{\tan \beta}{\tan \alpha}$| \nFrom the ray diagram for the eyepiece: $$\tan \beta = \frac{h'}{u_e}$$\nIf the final image is formed at the least distance of distinct vision $D$: $$\frac{1}{-D} - \frac{1}{-u_e} = \frac{1}{f_e} \implies \frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D}$$\nMultiply by $D$: $$\frac{D}{u_e} = 1 + \frac{D}{f_e}$$ \nFor the objective lens, the linear magnification is: $$m_o = \frac{h'}{h} = \frac{v_o}{u_o} \approx -\frac{v_o}{f_o}$$ \nThe total magnifying power is the product of the magnification of the objective and the eyepiece: $$m = m_o \times m_e = \left(-\frac{v_o}{u_o}\right) \left(1 + \frac{D}{f_e}\right)$$\nWhen the object is placed very close to the focus of the objective ($u_o \approx f_o$) and the image is formed near the eyepiece ($v_o \approx L$, tube length): $$m = -\frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$$\nThis is the required expression for the magnifying power.