CBSE 2025, Set 3

Question

(a) (i) Explain with the help of a labelled ray diagram the formation of final image by an astronomical telescope at infinity. Write the expression for its magnifying power.
(ii) The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. When the microscope is focussed on a certain object, the distance between the objective and eyepiece is observed to be $14	ext{ cm}$. Calculate the focal lengths of the objective and the eyepiece. (Given that the least distance of distinct vision = $25	ext{ cm}$)

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Astronomical Telescope (Normal Adjustment):** ![Astronomical Telescope](https://www.edvaya.com/images/cbse/physics/2025/set3/q32a-astronomical-telescope-infinity-answer.webp) * Light rays from a distant object enter the objective lens (large aperture and large focal length $f_o$) and form a real, inverted, and diminished image $A'B'$ at its principal focus. * This image $A'B'$ acts as a real object for the eyepiece lens. * For the final image to be formed at infinity, the intermediate image $A'B'$ must lie exactly at the principal focus ($f_e$) of the eyepiece. * The eyepiece acts as a simple magnifier, producing a highly magnified virtual final image at infinity. **Magnifying Power ($M$):** It is the ratio of the angle subtended at the eye by the image to the angle subtended by the object. $$\boxed{M = \frac{f_o}{f_e}}$$ **(ii) Compound Microscope Calculation:** Given: Total magnification $M = 20$ Magnification of eyepiece $m_e = 5$ Distance between lenses $L = 14	ext{ cm}$ Least distance of distinct vision $D = 25	ext{ cm}$ The total magnification is $M = m_o 	imes m_e$ $$20 = m_o 	imes 5 \Rightarrow m_o = 4$$ Since the final image is typically formed at the least distance of distinct vision: $$m_e = 1 + \frac{D}{f_e}$$ $$5 = 1 + \frac{25}{f_e} \Rightarrow 4 = \frac{25}{f_e}$$ $$\boxed{f_e = \frac{25}{4} = 6.25	ext{ cm}}$$ Now, for the eyepiece, $v_e = -25	ext{ cm}$. Using the lens formula: $$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$ $$\frac{1}{-25} - \frac{1}{u_e} = \frac{4}{25}$$ $$\frac{-1}{u_e} = \frac{4}{25} + \frac{1}{25} = \frac{5}{25} = \frac{1}{5}$$ $$u_e = -5	ext{ cm}$$ The distance between the lenses is $L = v_o + |u_e|$ $$14 = v_o + 5 \Rightarrow v_o = 9	ext{ cm}$$ For the objective lens, magnification $m_o = \frac{v_o}{|u_o|} = 4$ $$\frac{9}{|u_o|} = 4 \Rightarrow |u_o| = \frac{9}{4}	ext{ cm} = 2.25	ext{ cm}$$ So, $u_o = -\frac{9}{4}	ext{ cm}$. Using the lens formula for the objective: $$\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$$ $$\frac{1}{f_o} = \frac{1}{9} - \left(-\frac{4}{9}ight) = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}$$ $$\boxed{f_o = \frac{9}{5}	ext{ cm} = 1.8	ext{ cm}}$$

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