CBSE 2025, Set 1

Question

(b) (i) Write any two advantages of a compound microscope over a simple microscope. Draw a ray diagram for the image formation at the near point by a compound microscope and explain it.

(ii) A thin planoconcave lens with its curved face of radius of curvature $R$ is made of glass of refractive index $n_1$. It is placed coaxially in contact with a thin equiconvex lens of same radius of curvature of refractive index $n_2$. Obtain the power of the combination lens.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Compound microscope — advantages and ray diagram:** **Advantages of compound microscope over simple microscope (any two):** 1. **Larger magnification** — total magnification = $m_o 	imes m_e$, which is the product of two magnifications, much higher than a single lens. 2. **Brighter image** — the objective collects light from a small object and forms a bright, magnified real image for the eyepiece. 3. Any other valid advantage (e.g., better resolution, ability to study microorganisms). **Ray diagram — image at near point (D = 25 cm):** ![Compound microscope ray diagram for near point image](https://www.edvaya.com/images/cbse/physics/2025/set1/q33b-compound-microscope-near-point.webp) **Explanation:** - The **objective lens** (short focal length $f_o$) forms a real, inverted, magnified image $A'B'$ of the object $AB$. This image lies just within the focal length of the eyepiece. - The **eyepiece** (focal length $f_e$) acts as a simple microscope and forms a virtual, erect (with respect to $A'B'$, but inverted w.r.t. original object), magnified final image at the near point (25 cm from eye). - The total magnification: $M = m_o 	imes m_e = \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}ight)$ --- **(ii) Power of combined planoconcave + equiconvex lens:** **Planoconcave lens (refractive index $n_1$, radius of curved surface = $R$):** Using the lensmaker's equation $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}ight)$: For planoconcave lens: one flat face ($R_1 = \infty$) and one concave face ($R_2 = +R$, since for a planoconcave lens with curved surface as second surface): $$P_1 = \frac{1}{f_1} = (n_1 - 1)\left(\frac{1}{\infty} - \frac{1}{R}ight) = \frac{-(n_1 - 1)}{R}$$ **Equiconvex lens (refractive index $n_2$, both surfaces have radius $R$):** For equiconvex: $R_1 = +R$, $R_2 = -R$: $$P_2 = \frac{1}{f_2} = (n_2 - 1)\left(\frac{1}{R} - \frac{1}{-R}ight) = (n_2 - 1)\frac{2}{R}$$ **Combined power:** $$P = P_1 + P_2 = \frac{-(n_1 - 1)}{R} + \frac{2(n_2 - 1)}{R}$$ $$\boxed{P = \frac{2n_2 - n_1 - 1}{R}}$$ > **Sign convention check:** For the planoconcave lens, power is negative (diverging lens). For the equiconvex lens, power is positive. The net power depends on the relative values of $n_1$ and $n_2$.

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