Optics MCQs for NEET — Physics Questions with Answers

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.'

According to the NCERT (Eq. 9.43): 'The (linear) magnification due to the objective, namely $h'/h$, equals $L/f_o$.'

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.'

From NCERT Eq. 9.44(b): 'When the final image is formed at infinity, the angular magnification due to the eyepiece is $m_e = (D/f_e)$'.

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.

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.

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.

The formula for lenses in contact is $1/F = 1/f_1 + 1/f_2$, which simplifies to $F = f_1f_2 / (f_1 + f_2)$.

According to the NCERT text, 'Such big lenses tend to be very heavy and therefore, difficult to make and support by their edges. Further, it is rather difficult and expensive to make such large sized lenses which form images that are free from any kind of chromatic aberration and distortions.' The 'absence of image inversion' is not listed as a problem associated with the objective lens itself, but rather an issue that terrestrial telescopes address with additional inverting lenses. Refracting telescopes can produce inverted images, but this is a characteristic, not a problem inherent in large objective lenses.

The NCERT text states, 'In this case, the length of the telescope tube is $f_o + f_e$.' This refers to a refracting telescope when the final image is formed at infinity (normal adjustment).