CBSE 2023 · Region 5 · Set 1 · Q33 · 5 marks
(i)Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point.(ii)In a compound microscope an object is placed at a distance of $\displaystyle 1.5$ cm from the objective of focal length $\displaystyle 1.25$ cm . If the eye-piece has a focal length of $\displaystyle 5$ cm and the final image is formed at the near point, find the magnifying power of the microscope.(i)Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.(ii)The magnifying power of an astronomical telescope in normal adjustment is $\displaystyle 2.9$ and the objective and the eyepiece are separated by a distance of $\displaystyle 150$ cm . Find the focal lengths of the two lenses.
(i)
Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point.
(ii)
In a compound microscope an object is placed at a distance of $\displaystyle 1.5$ cm from the objective of focal length $\displaystyle 1.25$ cm . If the eye-piece has a focal length of $\displaystyle 5$ cm and the final image is formed at the near point, find the magnifying power of the microscope.
(i)
Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
(ii)
The magnifying power of an astronomical telescope in normal adjustment is $\displaystyle 2.9$ and the objective and the eyepiece are separated by a distance of $\displaystyle 150$ cm . Find the focal lengths of the two lenses.
Marking-scheme solution
(a) (ii)
Linear magnification due to the objective is
\[\begin{aligned}
& \tan \beta=\left(\frac{\mathrm{h}}{\mathrm{f}_{0}}\right)=\left(\frac{\mathrm{h}^{\prime}}{\mathrm{L}}\right) \\
& \mathrm{m}_{0}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{\mathrm{L}}{\mathrm{f}_{0}}
\end{aligned}
\]
Here L is the distance between the second focal point of the objective and the first focal point of the eyepiece.
Linear magnification due to eyepiece is
\[\mathrm{m}_{\mathrm{e}}=\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)
\]
Thus, the total magnification is, \(\displaystyle \mathrm{m}=\mathrm{m}_{0} \times \mathrm{m}_{\mathrm{e}}\)
\[\mathrm{m}=\frac{\mathrm{L}}{\mathrm{f}_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)
\]
\[\mathrm{m}=\frac{v_{0}}{u_{0}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)
\]Ray Optics and Optical InstrumentsOptical InstrumentsApplylong_answerhard
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.