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A compound microscope has objective lens of focal length $f_o = 2$ cm and eyepiece of focal length $f_e = 4$ cm. Length of the tube is $L = 40$ cm and least distance of distinct vision $D = 25$ cm. What is the magnification of the microscope?
Options
1
$-250$
2
$125$
3
$-125$
4
$250$
Correct Answer
125
Solution
1

Compound microscope magnification:

$$M = \frac{L}{f_o} \times \frac{D}{f_e}$$
2
$$M = \frac{40}{2} \times \frac{25}{4} = 20 \times 6.25 = \boxed{125}$$
$M = \dfrac{L}{f_o} \times \dfrac{D}{f_e} = \dfrac{40}{2} \times \dfrac{25}{4} = 125$
Theory: Optics
1. Compound Microscope

A compound microscope uses two converging lenses to achieve high magnification. Objective lens ($f_o$ short, a few mm): placed just outside its focal point. Creates a real, inverted, magnified intermediate image at distance $L$ from its back focal point. Eyepiece ($f_e$ longer): acts as a simple magnifying glass on the intermediate image. Total magnification: $M = m_o \times m_e = \dfrac{L}{f_o} \times \dfrac{D}{f_e}$ (image at near point, $D=25$ cm). For final image at infinity: $M = \dfrac{L}{f_o} \times \dfrac{D}{f_e}$ (same formula here).

2. Astronomical Telescope

Objective: large focal length $f_o$, large aperture (to collect more light from distant objects). Eyepiece: short focal length $f_e$ (for high magnification). Magnification: $M = f_o/f_e$ (for relaxed eye, final image at infinity). Length of telescope: $L = f_o + f_e$. Resolving power: $\theta_{min} = 1.22\lambda/D$ (Rayleigh criterion). Larger $D$ → better resolution. Reflecting telescope: uses concave mirror as objective (no chromatic aberration, easier to make large). Refracting: uses lens (chromatic aberration problem).

3. Simple Magnifier

Single convex lens used as magnifying glass. Object placed inside focal length. Virtual, erect, magnified image. Magnification: $M = 1 + D/f$ (image at near point $D=25$ cm). For relaxed eye: $M = D/f$. For this problem, eyepiece magnification $= D/f_e = 25/4 = 6.25$. Compound microscope achieves much higher magnification by multiplying: $125 = 20 \times 6.25$. Maximum useful magnification of optical microscope: $\sim 1500\times$ (limited by diffraction at visible wavelengths).

4. Lens Maker Equation

$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $n$ = refractive index of lens material. $R_1$, $R_2$ = radii of curvature of two surfaces (sign convention: positive if centre of curvature on transmission side). Thin lens formula: $1/v - 1/u = 1/f$. Power $P = 1/f$ (D). Lenses in contact: $P = P_1 + P_2$. For minimum spherical aberration: plano-convex lens oriented with flat side toward image.

5. Refraction at Spherical Surface

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ $n_1$ = refractive index of incident medium, $n_2$ = refractive index of refraction medium, $R$ = radius of curvature. Thin lens formula derived from applying this twice (at two surfaces). Refraction through prism: deviation $\delta = (\mu - 1)A$ for thin prism. Minimum deviation: $\sin[(A+\delta_m)/2] = \mu\sin(A/2)$.

6. Dispersion and Scattering

Dispersion: different wavelengths refract differently ($\mu$ varies with $\lambda$). Cauchy formula: $\mu = A + B/\lambda^2$. Violet bends most, red least (in glass). Rainbow: (1) refraction at entry + (2) TIR inside drop + (3) refraction at exit. Primary bow: $42°$, colours red outside-violet inside. Secondary bow: $51°$ (extra TIR), colours inverted. Scattering: Rayleigh scattering $\propto 1/\lambda^4$ → blue sky (blue scatters most), red/orange sunset (blue scattered away leaving red/orange).

7. Human Eye Defects

Myopia (short-sight): image forms in front of retina (eyeball too long or lens too powerful). Far point closer than infinity. Corrected by diverging lens (negative power). Hyperopia (long-sight): image behind retina. Near point farther than 25 cm. Corrected by converging lens. Astigmatism: different power in different meridians. Corrected by cylindrical lens. Presbyopia: loss of accommodation with age (lens hardens). Needs reading glasses (converging). Cataract: lens becomes opaque. Treated by lens replacement (IOL surgery).

8. Interference and Diffraction

Young double slit: $\beta = \lambda D/d$ (fringe width). Condition: coherent sources. Thin film interference: constructive if $2nt = (m+1/2)\lambda$ (phase change at one surface). Anti-reflection coating: $t = \lambda/(4n)$. Single slit diffraction: first minimum at $\sin\theta = \lambda/a$. Diffraction grating: $d\sin\theta = m\lambda$. Resolving power of grating $= mN$ ($N$ = number of slits). Limit of resolution (Rayleigh criterion): $\theta_{min} = 1.22\lambda/D$.

Frequently Asked Questions
1. Why is tube length L used in the formula instead of total length?
Tube length $L$ in compound microscope formula refers to the distance from the back focal point of objective to the front focal point of eyepiece (called the optical tube length). This is where the intermediate image forms. Total mechanical tube length = $L + f_o + f_e$. The formula $M = (L/f_o)(D/f_e)$ uses this optical tube length. Here $L = 40$ cm, $f_o = 2$ cm, $f_e = 4$ cm. The magnification $m_o = L/f_o$ gives the lateral magnification of the intermediate image by the objective.
2. How does increasing tube length affect magnification?
Magnification $M = (L/f_o)(D/f_e)$ is directly proportional to $L$. Doubling $L$ doubles magnification. However, longer tube lengths make microscopes larger and heavier. Modern microscopes use infinity-corrected optics: objective forms image at infinity, a tube lens then focuses it. This allows accessories (filters, cameras) to be inserted between objective and tube lens without affecting focus.
3. What is the resolving power of a microscope?
Resolving power = minimum separation between two points that can be distinguished. $d_{min} = 0.61\lambda/(n\sin\theta)$ = $0.61\lambda/NA$ where $NA = n\sin\theta$ is numerical aperture. To improve resolution: use shorter wavelength $\lambda$ (UV or electron microscopy), increase NA (use immersion oil, $n>1$). Electron microscope: uses electrons ($\lambda \sim 0.004$ nm) instead of light → 100,000x magnification. NEET considers only optical microscopes.
4. Why must object be placed just beyond focal length of objective?
For objective to form a real magnified intermediate image: object must be beyond $f_o$ (inside $f_o$ gives virtual image). For maximum magnification: object should be just slightly beyond $f_o$ (image forms far away, giving large magnification $m_o = v/u$ where $v >> u$). The tube length $L$ is the image distance beyond the objective back focal point. If $L = 40$ cm and $f_o = 2$ cm: $m_o = L/f_o = 40/2 = 20$. This means the intermediate image is 20x larger than the object.
5. What is the difference between magnification and resolving power?
Magnification: how much larger the image appears compared to the object (ratio of image size to object size). Can be increased by using lenses with smaller focal lengths. Resolving power: ability to distinguish two closely-spaced points as separate (limited by diffraction/wave nature of light). These are independent properties. A microscope with high magnification but poor resolving power gives a magnified but blurry image (called "empty magnification"). Useful magnification = 500-1000 × NA.
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