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For a compound microscope, the final image formed is always:
The NCERT states: 'Clearly, the final image is inverted with respect to the original object.'
What is the 'tube length' (L) in a compound microscope?
The NCERT defines it as: 'The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length $f_e$) is called the tube length of the compound microscope.'
The linear magnification due to the objective lens ($m_o$) of a compound microscope, represented by $h'/h$, is approximately given by:
According to the NCERT (Eq. 9.43): 'The (linear) magnification due to the objective, namely $h'/h$, equals $L/f_o$.'
For the total magnification of a compound microscope to be large, both the objective and eyepiece should ideally have:
The NCERT states: 'Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths.'
When the final image in a compound microscope is formed at infinity, the angular magnification due to the eyepiece ($m_e$) is given by:
From NCERT Eq. 9.44(b): 'When the final image is formed at infinity, the angular magnification due to the eyepiece is $m_e = (D/f_e)$'.
In a compound microscope, if the objective lens has a focal length $f_o = 1.0$ cm, and the eyepiece has a focal length $f_e = 2.0$ cm, with a tube length $L = 20$ cm, the total magnification (for image at infinity) will be approximately:
Using the formula $m = m_o m_e = (L/f_o)(D/f_e)$, and typical $D = 25$ cm: $m = (20/1.0)(25/2.0) = 20 imes 12.5 = 250$. This matches the example calculation in the NCERT.
Why is it difficult to make the focal length of lenses in a microscope much smaller than 1 cm?
While not explicitly stated as the only reason for the 1cm limit, the NCERT mentions: 'In practice, it is difficult to make the focal length much smaller than 1 cm. Also large lenses are required to make L large. ... In modern microscopes, multi-component lenses are used for both the objective and the eyepiece to improve image quality by minimising various optical aberrations (defects) in lenses.' The difficulty in minimizing focal length below 1 cm is related to managing aberrations and manufacturing practical, high-quality optics.
For best viewing through a compound microscope, why should the eye be positioned a short distance away from the eyepiece, rather than on it?
The NCERT (Q 9.25 (e)) implies this by asking 'Why? How much should be that short distance between the eye and eyepiece?'. The 'eye-ring' concept (exit pupil) is where all the rays from the object converge after passing through the eyepiece, and placing the eye there provides the largest field of view without vignetting.
What is the effective focal length of two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact?
The formula for lenses in contact is $1/F = 1/f_1 + 1/f_2$, which simplifies to $F = f_1f_2 / (f_1 + f_2)$.
Which of the following describes the fundamental nature of heat transfer by convection?
The NCERT text states, 'Convection is a mode of heat transfer by actual motion of matter.' This distinguishes it from conduction (molecular vibration) and radiation (electromagnetic waves).
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