- AWavelength of light.
- BDiameter of objective.
- CLength of the tube.
- DFocal length of eyepiece.
Explanation:
The resolving power of a telescope can be given as:
Resolving power $=\frac{1}{\text{d}(\theta)}=\frac{1}{1.22\lambda/ \text{D}}=\frac{\text{D}}{1.22}$ (wavelength)
So, resolving power can be increased by decreasing the wavelength and increasing the diameter of objective.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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(b) Being easy to handle |
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(c) Low cost |
(d) Quickness of observation |
Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are
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(a) Q > |
(b) |
(c) V > |
(d) |
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(a) |
(b) |
(c) |
(d) Zero |
A variable condenser is permanently connected to a 100 V battery. If the capacity is changed from 2μF to 10 μF, then change in energy is equal to
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(a) 2 |
(b) 2.5 |
(c) 3.5 |
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(a) E ∝ |
(b) E ∝ |
(c) E ∝ |
(d) E ∝ |
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(a) I0 |
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When is the real image formed?