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प्रश्न
A compound microscope has an objective of focal length 1.25 cm and eyepiece of focal length 5 cm. A small object is kept at 2.5 cm from the objective. If the final image formed is at infinity, find the distance between the objective and the eyepiece ?
उत्तर
Distance between the objective and the eyepiece, L =
\[u_0 = - 2 . 5 \text {cm and } f_0 = 1 . 25 cm\]
\[\text { Now }, - \frac{1}{u_0} + \frac{1}{v_0} = \frac{1}{f_0}\] \[or \ v_0 = 2 . 5 cm\]
To find ue, we have:
\[v_e = \infty\text { and } f_e = 5 cm\]
Calculating using the same formula as above, we get:
संबंधित प्रश्न
You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?
| Lenses | Power (D) | Aperture (cm) |
| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
Draw a ray diagram showing image formation in a compound microscope ?
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Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?
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The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye
How does the resolving power of a microscope change when
(i) the diameter of the objective lens is decreased?
(ii) the wavelength of the incident light is increased ?
Justify your answer in each case.
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